比较快也比较漂亮的熟练剖分模板(BZOJ 1036:树剖+线段树 BZOJ 1103:树剖+树状数组)
2017-09-08 23:47
453 查看
1036:
1103:
#include<bits/stdc++.h> #define N 100005 #define inf 1000000000 using namespace std; int n,q,a[4*N]; struct Edge{ int u,v,next; }G ; int tot=0,head ; int size[100005],wson[100005],fa[100005],d[100005],top[100005]; int tpos[100005],pre[100005],cnt=0; inline void addedge(int u,int v){ G[++tot].u=u;G[tot].v=v;G[tot].next=head[u];head[u]=tot; G[++tot].u=v;G[tot].v=u;G[tot].next=head[v];head[v]=tot; } void dfs1(int u,int f){ size[u]=1; for (int i=head[u];i;i=G[i].next){ int v=G[i].v;if (v==f)continue; d[v]=d[u]+1;fa[v]=u; dfs1(v,u); size[u]+=size[v]; if (size[v]>size[wson[u]])wson[u]=v; } } void dfs2(int u,int TP){ tpos[u]=++cnt;pre[cnt]=u;top[u]=TP; if (wson[u])dfs2(wson[u],TP); for (int i=head[u];i;i=G[i].next){ int v=G[i].v; if (v==fa[u]||v==wson[u])continue; dfs2(v,v); } } int sumv[4*N],maxv[4*N]; inline void pushup(int o){ sumv[o]=sumv[o*2]+sumv[o*2+1]; maxv[o]=max(maxv[o*2],maxv[o*2+1]); } void build(int o,int l,int r){ int mid=(l+r)/2; if (l==r){sumv[o]=maxv[o]=a[pre[l]];return;} build(o*2,l,mid);build(o*2+1,mid+1,r); pushup(o); } void update(int o,int l,int r,int q,int v){ int mid=(l+r)/2; if (l==r){sumv[o]=maxv[o]=v;return;} if (q<=mid)update(o*2,l,mid,q,v); else update(o*2+1,mid+1,r,q,v); pushup(o); } int querysum(int o,int l,int r,int ql,int qr){ int mid=(l+r)/2,ans=0; if (ql<=l&&r<=qr)return sumv[o]; if (ql<=mid)ans+=querysum(o*2,l,mid,ql,qr); if (qr>mid)ans+=querysum(o*2+1,mid+1,r,ql,qr); pushup(o); return ans; } int querymax(int o,int l,int r,int ql,int qr){ int mid=(l+r)/2,ans=-inf; if (ql<=l&&r<=qr)return maxv[o]; if (ql<=mid)ans=max(ans,querymax(o*2,l,mid,ql,qr)); if (qr>mid)ans=max(ans,querymax(o*2+1,mid+1,r,ql,qr)); pushup(o); return ans; } int qsum(int u,int v){ int ans=0; while (top[u]!=top[v]){ if (d[top[u]]<d[top[v]])swap(u,v); ans+=querysum(1,1,n,tpos[top[u]],tpos[u]); u=fa[top[u]]; } if (d[u]<d[v])swap(u,v); ans+=querysum(1,1,n,tpos[v],tpos[u]); return ans; } int qmax(int u,int v){ int ans=-inf; while (top[u]!=top[v]){ if (d[top[u]]<d[top[v]])swap(u,v); ans=max(ans,querymax(1,1,n,tpos[top[u]],tpos[u])); u=fa[top[u]]; } if (d[u]<d[v])swap(u,v); ans=max(ans,querymax(1,1,n,tpos[v],tpos[u])); return ans; } int main(){ memset(head,0,sizeof(head)); memset(a,0,sizeof(a)); scanf("%d",&n); for (int i=1;i<n;i++){ int u,v; scanf("%d%d",&u,&v); addedge(u,v); } for (int i=1;i<=n;i++)scanf("%d",&a[i]); d[1]=1;fa[1]=1;dfs1(1,-1);dfs2(1,1);build(1,1,n); scanf("%d",&q); while (q--){ int x,y; char s[10]; scanf("%s%d%d",s,&x,&y); if (s[1]=='H')update(1,1,n,tpos[x],y); if (s[1]=='M')printf("%d\n",qmax(x,y)); if (s[1]=='S')printf("%d\n",qsum(x,y)); } return 0; }
1103:
#include<bits/stdc++.h> using namespace std; const int INF = 0x3f3f3f3f; const int MAXN = 250000+100; struct Seg_Tree{ int sm[MAXN<<1]; inline int lowbit(int _x){ return _x&(-_x); } void build (int l,int r){ for (int i=l;i<=r;i++){ add(i,1); } } void add(int x,int val){ while (x<=MAXN){ sm[x]+=val; x+=lowbit(x); } } int sum(int x){ int res =0; while (x){ res+=sm[x]; x-=lowbit(x); } return res; } int query_sum(int l,int r){ return sum(r)-sum(l-1); } }tree; int first[MAXN*2]; int nxt[MAXN*2]; int des[MAXN*2]; int tpos[MAXN]; int deep[MAXN]; int top[MAXN]; int fa[MAXN]; int wson[MAXN]; int size[MAXN]; int n,q,tot=0,cnt=0; char s[10]; inline void add(int _u,int _v){ des[++tot] = _v; nxt[tot] = first[_u]; first[_u] = tot; } void input(){ scanf("%d",&n); for (int i=1;i<n;i++){ int u,v; scanf("%d%d",&u,&v); add(u,v); add(v,u); } } void dfs(int node,int father){ deep[node] = deep[father]+1; fa[node] = father; size[node] =1; for (int t = first[node];t;t = nxt[t]){ int v = des[t]; if (v==father){ continue; } dfs(v,node); if (size[v]>size[wson[node]]){ wson[node] = v; } size[node]+=size[v]; } } void dfs2(int node,int father,int chain){ top[node] = chain; tpos[node] = ++cnt; if (wson[node]){ dfs2(wson[node],node,chain); } for (int t = first[node];t;t = nxt[t]){ int v = des[t]; if (v==father||v ==wson[node]){ continue; } dfs2(v,node,v); } } void init(){ dfs(1,0); dfs2(1,0,1); tree.build(2,n); } int get_sum(int u,int v){ int res =0; while (top[u]!=top[v]){ if (deep[top[u]]<deep[top[v]]){ swap(u,v); } res+= tree.query_sum(tpos[top[u]],tpos[u]); u = fa[top[u]]; } if (deep[u]<deep[v]){ swap(u,v); } res += tree.query_sum(tpos[v],tpos[u]); return res; } void modify(int u,int v){ if (fa[u]!=v){ swap(u,v); } tree.add(tpos[u],-1); } void solve(){ scanf("%d",&q); q+=n-1; while (q--){ scanf("%s",s); if (s[0]=='W'){ int x; scanf("%d",&x); printf("%d\n",get_sum(1,x)); }else{ int x,y; scanf("%d%d",&x,&y); modify(x,y); } } } int main(){ input(); init(); solve(); return 0; }
相关文章推荐
- BZOJ-1036 树的统计Count 链剖线段树(模板)=(树链剖分+线段树)
- BZOJ-1036 树的统计Count 链剖线段树(模板)=(树链剖分+线段树)
- BZOJ 2758 Blinker的噩梦(扫描线+熟练剖分+树状数组)
- BZoj 1036: [ZJOI2008]树的统计Count【树链剖分+线段树--模板题】
- [POI2007][BZOJ1103] 大都市meg|dfs序|树状数组
- 【BZOJ3110】【整体二分+树状数组区间修改/线段树】K大数查询
- bzoj 1036 [ZJOI2008]树的统计Count (树链剖分 + 线段树)
- BZOJ[1036] [ZJOI2008]树的统计Count 树链剖分模板
- [置顶] bzoj 1036 树的统计Count 点权值模板
- [Bzoj1103][POI2007]大都市meg (dfs序+ 树状数组)
- BZOJ-1036: [ZJOI2008]树的统计Count (树链剖分 线段树 单点修改 区间查询 入门题)
- [BZOJ4785][ZJOI2017]树状数组(概率+二维线段树)
- 【bzoj1103】大都市meg 树链剖分+线段树
- hdu 1166 敌兵布阵 (线段树、树状数组模板,单点更新)
- 线段树和树状数组的全面配合与比较
- BZOJ 1036 [ZJOI2008]树的统计Count | 树链剖分模板
- 【树链剖分+线段树】BZOJ1036 [ZJOI2008]树的统计Count
- bzoj1036 树链剖分模板
- bzoj 1036 [ZJOI2008] 树的统计 树链剖分模板题
- bzoj1036 树的统计Count 【树链剖分+线段树】