/**************************************************************Problem: 1001User: whymheLanguage: C++Result: AcceptedTime:1236 msMemory:106760 kb ****************************************************************///Pro:P4001 [BJOI2006]狼抓兔子//大概就是求個(gè)最小割 //這個(gè)題的唯一的難點(diǎn)大概就是如何給網(wǎng)格圖編號(hào) //一個(gè)比較坑的地方： //邊都是雙向的，所以正反向弧的容量都是flow #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<queue> using namespace std;inline int read() {char c=getchar();int num=0;for(;!isdigit(c);c=getchar());for(;isdigit(c);c=getchar())num=num*10+c-'0';return num; }const int N=1e6+5;int n,m,S,T; int head[N],num_edge; struct Edge {int v,flow,nxt; }edge[N<<3];inline void add_edge(int u,int v,int flow) {edge[++num_edge].v=v;edge[num_edge].flow=flow;edge[num_edge].nxt=head[u];head[u]=num_edge; }int from[N],dep[N]; inline bool bfs() {queue<int> que;int now;for(now=S;now<=T;++now)dep[now]=0,from[now]=head[now];que.push(S),dep[S]=1;while(!que.empty()){now=que.front(),que.pop();for(int i=head[now],v;i;i=edge[i].nxt){v=edge[i].v;if(dep[v]||!edge[i].flow)continue;dep[v]=dep[now]+1;if(v==T)return 1;que.push(v);}}return 0; }int dfs(int u,int flow) {if(u==T||!flow)return flow;int outflow=0,tmp;for(int &i=from[u],v;i;i=edge[i].nxt){v=edge[i].v;if(dep[v]!=dep[u]+1)continue;tmp=dfs(v,min(flow,edge[i].flow));if(!tmp)continue;flow-=tmp;outflow+=tmp;edge[i].flow-=tmp;edge[i^1].flow+=tmp;if(!flow)return outflow;}dep[u]=0;return outflow; }int main() {num_edge=1;n=read(),m=read();S=1,T=n*m;for(int i=1,p,f;i<=n;++i){for(int j=1;j<m;++j){p=(i-1)*m+j;f=read();add_edge(p,p+1,f);add_edge(p+1,p,f);}}for(int i=1,p,f;i<n;++i){for(int j=1;j<=m;++j){p=(i-1)*m+j;f=read(),add_edge(p,p+m,f);add_edge(p+m,p,f);}}for(int i=1,p,f;i<n;++i){for(int j=1;j<m;++j){p=(i-1)*m+j;f=read();add_edge(p,p+m+1,f);add_edge(p+m+1,p,f);}}long long ans=0;while(bfs())ans+=1ll*dfs(S,0x7fffffff);printf("%lld",ans);return 0; }

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轉(zhuǎn)載于:https://www.cnblogs.com/lovewhy/p/9633649.html

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