hdu 5044 树链剖分(点更新、边更新的更优美姿势才能过)
2015-09-02 14:18
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Link:http://acm.hdu.edu.cn/showproblem.php?pid=5044
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2673 Accepted Submission(s): 467
Problem Description
You are given a tree (an acyclic undirected connected graph) with N nodes. The tree nodes are numbered from 1 to N
There are N - 1 edges numbered from 1 to N - 1.
Each node has a value and each edge has a value. The initial value is 0.
There are two kind of operation as follows:
● ADD1 u v k: for nodes on the path from u to v, the value of these nodes increase by k.
● ADD2 u v k: for edges on the path from u to v, the value of these edges increase by k.
After finished M operation on the tree, please output the value of each node and edge.
Input
The first line of the input is T (1 ≤ T ≤ 20), which stands for the number of test cases you need to solve.
The first line of each case contains two integers N ,M (1 ≤ N, M ≤105),denoting the number of nodes and operations, respectively.
The next N - 1 lines, each lines contains two integers u, v(1 ≤ u, v ≤ N ), denote there is an edge between u,v and its initial value is 0.
For the next M line, contain instructions “ADD1 u v k” or “ADD2 u v k”. (1 ≤ u, v ≤ N, -105 ≤ k ≤ 105)
Output
For each test case, print a line “Case #t:”(without quotes, t means the index of the test case) at the beginning.
The second line contains N integer which means the value of each node.
The third line contains N - 1 integer which means the value of each edge according to the input order.
Sample Input
Sample Output
Source
2014 ACM/ICPC Asia Regional Shanghai Online
题意:给出一棵树,有两种操作:1 给路径上的所有点加vi,2所有边加vi。最后输出所有点的权值和所有边的权值。
编程思想:要换另一种姿势的树链剖分才能过。。
这种姿势的代码参考自:http://blog.csdn.net/qinzhenhua100/article/details/39716851
AC code:
附上我自己写的超时代码(以供对比反思借鉴):
TLE code:
Tree
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2673 Accepted Submission(s): 467
Problem Description
You are given a tree (an acyclic undirected connected graph) with N nodes. The tree nodes are numbered from 1 to N
There are N - 1 edges numbered from 1 to N - 1.
Each node has a value and each edge has a value. The initial value is 0.
There are two kind of operation as follows:
● ADD1 u v k: for nodes on the path from u to v, the value of these nodes increase by k.
● ADD2 u v k: for edges on the path from u to v, the value of these edges increase by k.
After finished M operation on the tree, please output the value of each node and edge.
Input
The first line of the input is T (1 ≤ T ≤ 20), which stands for the number of test cases you need to solve.
The first line of each case contains two integers N ,M (1 ≤ N, M ≤105),denoting the number of nodes and operations, respectively.
The next N - 1 lines, each lines contains two integers u, v(1 ≤ u, v ≤ N ), denote there is an edge between u,v and its initial value is 0.
For the next M line, contain instructions “ADD1 u v k” or “ADD2 u v k”. (1 ≤ u, v ≤ N, -105 ≤ k ≤ 105)
Output
For each test case, print a line “Case #t:”(without quotes, t means the index of the test case) at the beginning.
The second line contains N integer which means the value of each node.
The third line contains N - 1 integer which means the value of each edge according to the input order.
Sample Input
2 4 2 1 2 2 3 2 4 ADD1 1 4 1 ADD2 3 4 2 4 2 1 2 2 3 1 4 ADD1 1 4 5 ADD2 3 2 4
Sample Output
Case #1: 1 1 0 1 0 2 2 Case #2: 5 0 0 5 0 4 0
Source
2014 ACM/ICPC Asia Regional Shanghai Online
题意:给出一棵树,有两种操作:1 给路径上的所有点加vi,2所有边加vi。最后输出所有点的权值和所有边的权值。
编程思想:要换另一种姿势的树链剖分才能过。。
这种姿势的代码参考自:http://blog.csdn.net/qinzhenhua100/article/details/39716851
AC code:
#pragma comment(linker, "/STACK:102400000,102400000") #include<stdio.h> #include<iostream> #include<string.h> using namespace std; #define N 100010 struct pp { int u,v; }ed ; struct node { int u,v,next; }bian[N*2]; int e,id,dep ,son ,father ,sz ,ti ,mark1 ,mark2 ,top ,head ; __int64 a ,b ,ans1 ,ans2 ; void add(int u,int v) { bian[e].u=u; bian[e].v=v; bian[e].next=head[u]; head[u]=e++; } void dfs1(int u,int fa) { int i,v; dep[u]=dep[fa]+1; son[u]=0; father[u]=fa; sz[u]=1; for(i=head[u];i!=-1;i=bian[i].next) { v=bian[i].v; if(v==fa) continue; dfs1(v,u); sz[u]+=sz[v]; if(sz[son[u]]<sz[v]) son[u]=v; } } void dfs2(int u,int fa) { int i,v; ti[u]=id++; mark1[id-1]=u; top[u]=fa; if(son[u]!=0) dfs2(son[u],fa); for(i=head[u];i!=-1;i=bian[i].next) { v=bian[i].v; if(v==father[u]||v==son[u]) continue; dfs2(v,v); } } void getnode(int u,int v,int k) { while(top[u]!=top[v]) { if(dep[top[u]]>dep[top[v]]) swap(u,v); a[ti[top[v]]]+=k; a[ti[v]+1]-=k; v=father[top[v]]; } if(ti[u]>ti[v]) swap(u,v); a[ti[u]]+=k; a[ti[v]+1]-=k; } void getedge(int u,int v,int k) { while(top[u]!=top[v]) { if(dep[top[u]]>dep[top[v]]) swap(u,v); b[ti[top[v]]]+=k; b[ti[v]+1]-=k; v=father[top[v]]; } if(ti[u]>ti[v]) swap(u,v); if(u!=v) { b[ti[u]+1]+=k; b[ti[v]+1]-=k; } } int main() { int t,cnt=1,n,m,i,u,v,k; __int64 s; char str[10]; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); memset(a,0,sizeof(a)); memset(head,-1,sizeof(head)); memset(b,0,sizeof(b)); e=0; for(i=1;i<n;i++) { scanf("%d%d",&ed[i].u,&ed[i].v); add(ed[i].u,ed[i].v); add(ed[i].v,ed[i].u); } sz[0]=0; id=1; dep[1]=0; dfs1(1,1); dfs2(1,1); for(i=1;i<=m;i++) { scanf("%s%d%d%d",str,&u,&v,&k); if(strcmp(str,"ADD1")==0) getnode(u,v,k); else getedge(u,v,k); } for(i=1;i<n;i++) { if(dep[ed[i].u]<dep[ed[i].v]) mark2[ti[ed[i].v]]=i; else mark2[ti[ed[i].u]]=i; } printf("Case #%d:\n",cnt++); s=0; for(i=1;i<=n;i++) { s+=a[i]; ans1[mark1[i]]=s; } for(i=1;i<=n;i++) { if(i==1) printf("%I64d",ans1[i]); else printf(" %I64d",ans1[i]); } printf("\n"); s=0; for(i=2;i<=n;i++) { s+=b[i]; ans2[mark2[i]]=s; } for(i=1;i<n;i++) { if(i==1) printf("%I64d",ans2[i]); else printf(" %I64d",ans2[i]); } printf("\n"); } return 0; }
附上我自己写的超时代码(以供对比反思借鉴):
TLE code:
#pragma comment(linker, "/STACK:1024000000,1024000000") //因OJ采用Windows系统,要加入这一行用于 进行手动扩栈,这样就不会引起爆栈 #include<stdio.h> #include<string.h> #include<vector> const int N=100010; using namespace std; int head , to[N << 1], next1[N << 1], tot;//边信息 int top ; //top[v]=u表示点v,u在一个链中,且u是这个链深度最小的点(即顶端) int fath ; //记录父节点 int deep ; //每个点在树上的深度 int num ; //每棵子树的节点个数 int son ; //选的重边子节点 int p ; //树上每个点在线段树中所对应的点 int pos; //线段树叶子结点总数 void addEdge(const int& u, const int& v) { to[tot] = v, next1[tot] = head[u], head[u] = tot++; } void addUndirEdge(const int& u, const int& v) { addEdge(u, v), addEdge(v, u); } void init(int n) { pos=0; tot=0; memset(son,-1,sizeof(son)); memset(head,-1,sizeof(head)); } void dfs1(int u,int pre,int d)//第一遍dfs求出fath,deep,num,son { deep[u]=d; fath[u]=pre; num[u]=1; for (int i = head[u]; i != -1; i = next1[i]) { int v = to[i]; if(v==pre)continue; dfs1(v,u,d+1); num[u]+=num[v]; if(son[u]==-1||num[v]>num[son[u]]) son[u]=v; } } void getpos(int u,int root) { top[u]=root; p[u]=++pos;//从1开始 if(son[u]==-1) return ; getpos(son[u],root); for (int i = head[u]; i != -1; i = next1[i]) { int v = to[i]; if(son[u]!=v&&v!=fath[u]) getpos(v,v); } } //线段树 struct tree { int sum,maxv,toc,addc;//区间和,区间最大值, }root[N*4],root2[N*2]; int val ,val2 ;//权值 int MAX(int a,int b) { return a>b?a:b; } void build(int l,int r,int k)//建树,范围为l~r,k为根节点的线段树。 build(1,pos,1); { int mid=(l+r)/2; root[k].addc=0; root[k].toc=0; if(l==r){ root[k].sum=root[k].maxv=val[l]; return ; } build(l,mid,k<<1); build(mid+1,r,k<<1|1); root[k].sum=root[k<<1].sum+root[k<<1|1].sum; root[k].maxv=MAX(root[k<<1].maxv,root[k<<1|1].maxv); } void build2(int l,int r,int k)//建树,范围为l~r,k为根节点的线段树。 build(1,pos,1); { int mid=(l+r)/2; root2[k].addc=0; root2[k].toc=0; if(l==r){ root2[k].sum=root2[k].maxv=val2[l]; return ; } build2(l,mid,k<<1); build2(mid+1,r,k<<1|1); root2[k].sum=root2[k<<1].sum+root2[k<<1|1].sum; root2[k].maxv=MAX(root2[k<<1].maxv,root2[k<<1|1].maxv); } void upson(int k,int l,int r)//更新儿子 { int mid=(l+r)/2; if(root[k].toc) { root[k<<1].sum=(mid-l+1)*root[k].toc; root[k<<1].maxv=root[k].toc; root[k<<1].toc=root[k].toc; root[k<<1].addc=0; root[k<<1|1].sum=(r-mid)*root[k].toc; root[k<<1|1].maxv=root[k].toc; root[k<<1|1].toc=root[k].toc; root[k<<1|1].addc=0; root[k].toc=0; } if(root[k].addc) { root[k<<1].sum+=(mid-l+1)*root[k].addc; root[k<<1].maxv+=root[k].addc; root[k<<1].addc+=root[k].addc; root[k<<1|1].sum+=(r-mid)*root[k].addc; root[k<<1|1].maxv+=root[k].addc; root[k<<1|1].addc+=root[k].addc; root[k].addc=0; } } void upson2(int k,int l,int r)//更新儿子 { int mid=(l+r)/2; if(root2[k].toc) { root2[k<<1].sum=(mid-l+1)*root2[k].toc; root2[k<<1].maxv=root2[k].toc; root2[k<<1].toc=root2[k].toc; root2[k<<1].addc=0; root2[k<<1|1].sum=(r-mid)*root2[k].toc; root2[k<<1|1].maxv=root2[k].toc; root2[k<<1|1].toc=root2[k].toc; root2[k<<1|1].addc=0; root2[k].toc=0; } if(root2[k].addc) { root2[k<<1].sum+=(mid-l+1)*root2[k].addc; root2[k<<1].maxv+=root2[k].addc; root2[k<<1].addc+=root2[k].addc; root2[k<<1|1].sum+=(r-mid)*root2[k].addc; root2[k<<1|1].maxv+=root2[k].addc; root2[k<<1|1].addc+=root2[k].addc; root2[k].addc=0; } } void updata1(int l,int r,int k,const int L,const int R,int c)//从范围为1(l)~pos(r),根节点为1(k)的线段树中更新 成段更新的区间位置为(L,R)的元素值为c { if(L<=l&&r<=R) { root[k].sum=(r-l+1)*c; root[k].maxv=c; root[k].toc=c; root[k].addc=0; return ; } int mid=(l+r)/2; upson(k,l,r); if(L<=mid) updata1(l,mid,k<<1,L,R,c); if(mid<R) updata1(mid+1,r,k<<1|1,L,R,c); root[k].sum=root[k<<1].sum+root[k<<1|1].sum; root[k].maxv=MAX(root[k<<1].maxv,root[k<<1|1].maxv); } void updata12(int l,int r,int k,const int L,const int R,int c)//从范围为1(l)~pos(r),根节点为1(k)的线段树中更新 成段更新的区间位置为(L,R)的元素值为c { if(L<=l&&r<=R) { root2[k].sum=(r-l+1)*c; root2[k].maxv=c; root2[k].toc=c; root2[k].addc=0; return ; } int mid=(l+r)/2; upson2(k,l,r); if(L<=mid) updata12(l,mid,k<<1,L,R,c); if(mid<R) updata12(mid+1,r,k<<1|1,L,R,c); root2[k].sum=root2[k<<1].sum+root2[k<<1|1].sum; root2[k].maxv=MAX(root2[k<<1].maxv,root2[k<<1|1].maxv); } void updata2(int l,int r,int k, int L, int R,int c)//从范围为1(l)~pos(r),根节点为1(k)的线段树中更新 成段更新的区间位置为(L,R)的元素值为原来的值加上增量c { if(L<=l&&r<=R) { root[k].sum+=(r-l+1)*c; root[k].maxv+=c; root[k].addc+=c; return ; } int mid=(l+r)/2; upson(k,l,r); if(L<=mid) updata2(l,mid,k<<1,L,R,c); if(mid<R) updata2(mid+1,r,k<<1|1,L,R,c); root[k].sum=root[k<<1].sum+root[k<<1|1].sum; root[k].maxv=MAX(root[k<<1].maxv,root[k<<1|1].maxv); } void updata22(int l,int r,int k, int L, int R,int c)//从范围为1(l)~pos(r),根节点为1(k)的线段树中更新 成段更新的区间位置为(L,R)的元素值为原来的值加上增量c { if(L<=l&&r<=R) { root2[k].sum+=(r-l+1)*c; root2[k].maxv+=c; root2[k].addc+=c; return ; } int mid=(l+r)/2; upson2(k,l,r); if(L<=mid) updata22(l,mid,k<<1,L,R,c); if(mid<R) updata22(mid+1,r,k<<1|1,L,R,c); root2[k].sum=root2[k<<1].sum+root2[k<<1|1].sum; root2[k].maxv=MAX(root2[k<<1].maxv,root2[k<<1|1].maxv); } int sum,maxv; void query(int l,int r,int k,int L,int R)//查询范围为1(l)~pos(r),根节点为1(k)的线段树[L,R]区间的和(全局变量sum)与最大值(全局变量maxv) { if(L<=l&&r<=R) { sum+=root[k].sum; maxv=MAX(maxv,root[k].maxv); return ; } int mid=(l+r)/2; upson(k,l,r); if(L<=mid) query(l,mid,k<<1,L,R); if(mid<R) query(mid+1,r,k<<1|1,L,R); } void query2(int l,int r,int k,int L,int R)//查询范围为1(l)~pos(r),根节点为1(k)的线段树[L,R]区间的和(全局变量sum)与最大值(全局变量maxv) { if(L<=l&&r<=R) { sum+=root2[k].sum; maxv=MAX(maxv,root2[k].maxv); return ; } int mid=(l+r)/2; upson2(k,l,r); if(L<=mid) query2(l,mid,k<<1,L,R); if(mid<R) query2(mid+1,r,k<<1|1,L,R); } void swp(int &a,int &b)//交换a、b { int tt; tt=a; a=b; b=tt; } void Operat0(int u,int v) //查询u->v路径上节点的权值的和全局变量sum)、u->v路径上节点的最大权值(全局变量maxv) { int f1=top[u], f2=top[v]; sum=0; maxv=0; while(f1!=f2) { if(deep[f1]<deep[f2]) { swp(f1,f2); swp(u,v); } query(1,pos,1,p[f1],p[u]);//对应查询范围为1(l)~pos(r),根节点为1(k)的线段树[L,R]区间的和(全局变量sum)与最大值(全局变量maxv) u=fath[f1]; f1=top[u]; } if(deep[u]>deep[v]) swp(u,v); query(1,pos,1,p[u],p[v]);//点权查询与边权查询的区别之一:点p[u],边p[son[u]] } void Operat02(int u,int v) //查询u->v路径上节点的权值的和全局变量sum)、u->v路径上节点的最大权值(全局变量maxv) { int f1=top[u], f2=top[v]; sum=0; maxv=0; while(f1!=f2) { if(deep[f1]<deep[f2]) { swp(f1,f2); swp(u,v); } query2(1,pos,1,p[f1],p[u]);//对应查询范围为1(l)~pos(r),根节点为1(k)的线段树[L,R]区间的和(全局变量sum)与最大值(全局变量maxv) u=fath[f1]; f1=top[u]; } if(deep[u]>deep[v]) swp(u,v); query2(1,pos,1,p[u],p[v]);//点权查询与边权查询的区别之一:点p[u],边p[son[u]] } void Operat1(int u,int v,int c)//表示从u点到v点的路径的每条边权都变成c { int f1=top[u], f2=top[v]; while(f1!=f2) { if(deep[f1]<deep[f2]) { swp(f1,f2); swp(u,v); } updata1(1,pos,1,p[f1],p[u],c);//对应从范围为1(l)~pos(r),根节点为1(k)的线段树中更新 成段更新的区间位置为(p[f1],p[u])的元素值为c u=fath[f1]; f1=top[u]; } if(deep[u]>deep[v]) swp(u,v); updata1(1,pos,1,p[u],p[v],c);//点权更新与边权更新的区别之一:点p[u],边p[son[u]] } void Operat2(int u,int v,int c)//表示从u点到v点的路径的每个点都加上c { int f1=top[u], f2=top[v]; while(f1!=f2) { if(deep[f1]<deep[f2]) { swp(f1,f2); swp(u,v); } updata2(1,pos,1,p[f1],p[u],c);//对应从范围为1(l)~pos(r),根节点为1(k)的线段树中更新 成段更新的区间位置为(p[f1],p[u])的元素值为原来的值加上增量c u=fath[f1]; f1=top[u];//向上迭代 } if(deep[u]>deep[v]) swp(u,v); updata2(1,pos,1,p[u],p[v],c);//点权更新与边权更新的区别之一:点p[u],边p[son[u]] } void Operat22(int u,int v,int c)//表示从u点到v点的路径的每个点都加上c { int f1=top[u], f2=top[v]; while(f1!=f2) { if(deep[f1]<deep[f2]) { swp(f1,f2); swp(u,v); } updata22(1,pos,1,p[f1],p[u],c);//对应从范围为1(l)~pos(r),根节点为1(k)的线段树中更新 成段更新的区间位置为(p[f1],p[u])的元素值为原来的值加上增量c u=fath[f1]; f1=top[u];//向上迭代 } if(u==v) return ;//点权更新与边权更新的区别之一:若是点权更新,则这一句要注释掉!!! if(deep[u]>deep[v]) swp(u,v); updata22(1,pos,1,p[son[u]],p[v],c);//边权更新用这句,若是点权更新则用下一句代码 } struct EDG { int u,v,c; }edg ; int point ;//点权值 char ope[10]; void scanf ( int& x , char c = 0 , int flag = 0 ) { while ( ( c = getchar () ) != '-' && ( c < '0' || c > '9' ) ) ; if ( c == '-' ) flag = 1 , x = 0 ; else x = c - '0' ; while ( ( c = getchar () ) >= '0' && c <= '9' ) x = x * 10 + c - '0' ; if ( flag ) x = -x ; } inline void out(int num){ bool flag=false; if(num<0){ putchar('-'); num=-num; } int ans[22],top=0; while(num!=0){ ans[top++]=num%10; num/=10; } if(top==0) putchar('0'); for(int i=top-1;i>=0;i--){ char ch=ans[i]+'0'; putchar(ch); } } int main() { int cas,T; int n,m,q,op,a,b,c; scanf(cas); T=0; while(cas--) { T++; scanf(n); scanf(q); //scanf("%d%d",&n,&q); init(n);//初始化 m=n-1; for(int i=1;i<=m;i++) { scanf(edg[i].u); scanf(edg[i].v); //scanf("%d%d",&edg[i].u,&edg[i].v); addUndirEdge(edg[i].u, edg[i].v); edg[i].c=0; point[i]=0; } point =0; dfs1(1,1,1);//第一遍dfs求出fath,deep,num,son getpos(1,1); pos=n; //线段树叶子结点总数 for(int i=1;i<n;i++)//将边的权值录入对应的线段树的位置上 { if(deep[edg[i].u]>deep[edg[i].v]) edg[i].v=edg[i].u; val2[p[edg[i].v]]=edg[i].c; //转换成在线段树上的对应位置 } for(int i=1;i<=n;i++) { val[p[i]]=point[i];//一定要注意转换成在线段树上的对应位置 } build(1,pos,1); build2(1,pos,1); while(q--) { scanf("%s",&ope); scanf(a),scanf(b),scanf(c); if(ope[3]=='1') { Operat2(a,b,c);//表示从a点到b点的路径的每条边权都加上c } else { Operat22(a,b,c);//表示从a点到b点的路径的每条边权都加上c } } printf("Case #%d:\n",T); Operat0(1,1); printf("%d",sum); for(int i=2;i<=n;i++) { Operat0(i,i); printf(" %d",sum); } puts(""); Operat02(edg[1].u, edg[1].v); //printf("%d",sum); out(sum); for(int i=2;i<=m;i++) { Operat02(edg[i].u, edg[i].v); // printf(" %d",sum); printf(" "); out(sum); } puts(""); } }
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