【网络流】hdu3046 Pleasant sheep and big big wolf
2014-09-17 18:13
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题意:青青大草原上有喜羊羊他们的小羊们还有灰太狼一帮狼,想再要加栅栏使得每只羊和每只狼隔开,求需要的最少的栅栏数量。
难度:2
题解:1连源点,2连汇点,容量无限,各个格子间连边容量为1,求最小割(题解转载自notonlysuccess.com)
hdu3998
题意:(同类问题应该来自《线性规划与网络流》24题)同类问题共三问:(1)最长上升子序列长度(2)每个数用一遍,能够得到的最多长度为s的子序列个数(3)同第二问,a[1]与a
可用多次
难度:5
题解:拆点后做最大流。第一问DP求出s,并得到每个数的f[i]。新增源点s,汇点t,对于节点i,若f[i]=1,则与源点连一条容量为1的弧,若f[i]=s,则与汇点连一条容量为1的弧,每个点与拆点后所得到的点连一条容量为1的弧,入度与原节点连,出度与拆点后的点连,若(j>i) (a[j]>a[i]) (f[j]=f[i]+1),则I j 之间连一条容量为1的弧。第三问将1与s的容量,1与1’的容量,n与n’的容量修改为+oo,若f
=s,则n’与t的容量也为+oo 再算一遍最大流。
难度:2
题解:1连源点,2连汇点,容量无限,各个格子间连边容量为1,求最小割(题解转载自notonlysuccess.com)
hdu3998
题意:(同类问题应该来自《线性规划与网络流》24题)同类问题共三问:(1)最长上升子序列长度(2)每个数用一遍,能够得到的最多长度为s的子序列个数(3)同第二问,a[1]与a
可用多次
难度:5
题解:拆点后做最大流。第一问DP求出s,并得到每个数的f[i]。新增源点s,汇点t,对于节点i,若f[i]=1,则与源点连一条容量为1的弧,若f[i]=s,则与汇点连一条容量为1的弧,每个点与拆点后所得到的点连一条容量为1的弧,入度与原节点连,出度与拆点后的点连,若(j>i) (a[j]>a[i]) (f[j]=f[i]+1),则I j 之间连一条容量为1的弧。第三问将1与s的容量,1与1’的容量,n与n’的容量修改为+oo,若f
=s,则n’与t的容量也为+oo 再算一遍最大流。
#include<cstdio> #include<cstring> #include<iostream> #define maxn 205000 #define maxm 2050000 #define inf 2000000000 using namespace std; int gap[maxn],dis[maxn],pre[maxn],cur[maxn]; int n,NV,m; struct Edge { int v,val; int next; Edge(){} Edge( int V , int NEXT , int W = 0 ):v(V),next(NEXT),val(W){} }edge[maxm]; int maxflow; int cnt_edge,head[maxn]; void addedge( int u , int v , int flow = 0 ) { edge[cnt_edge] = Edge(v,head[u],flow); head[u] = cnt_edge++; edge[cnt_edge] = Edge(u,head[v]); head[v] = cnt_edge++; } void init() { cnt_edge = 0; memset(head,-1,sizeof(int)*(NV+1)); } int Sap( int st, int en ) { memset(dis,0,sizeof(int)*( NV+1)); memset(gap,0,sizeof(int)*( NV+1)); for( int i = 0 ; i < NV ; i++ ) cur[i] = head[i]; int u = pre[st] = st,maxflow = 0,aug = inf; gap[0] = NV; while( dis[st] < NV ) { loop: for( int &i = cur[u]; i != -1 ; i = edge[i].next ) { int v =edge[i].v; if( edge[i].val && dis[u] == dis[v]+1) { aug = aug < edge[i].val? aug: edge[i].val; pre[v] = u; u = v; if( v == en ) { maxflow += aug; for( u = pre[u]; v != st ; v = u,u = pre[u] ) { edge[cur[u]].val -= aug; edge[cur[u]^1].val += aug; } aug = inf; } goto loop; } } int mindis = NV; for( int i = head[u]; i != -1 ; i = edge[i].next ) { int v = edge[i].v; if( edge[i].val && mindis > dis[v] ) { cur[u] = i; mindis = dis[v]; } } if( --gap[dis[u]] == 0 ) break; gap[ dis[u] = mindis+1 ]++; u = pre[u]; } return maxflow; } int N,M; int main() { int cas = 1; while(~scanf("%d%d",&N,&M)) { printf("Case %d:\n",cas++); int st = N*M , en = st + 1; NV = en + 1; init(); for(int i=0;i<N;i++) for(int j=0;j<M;j++) { int ff; scanf("%d",&ff); if(ff == 1) addedge(st,i*M+j,inf); if(ff == 2) addedge(i*M+j,en,inf); if(i<N-1) addedge(i*M+j,i*M+j+M,1); if(i>0) addedge(i*M+j,i*M+j-M,1); if(j<M-1) addedge(i*M+j,i*M+j+1,1); if(j > 0) addedge(i*M+j,i*M+j-1,1); } int ans = Sap(st,en); printf("%d\n",ans); } return 0; }
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