HDU 3046——Pleasant sheep and big big wolf

網路流Dinic演算法

狼為源點，羊為匯點。

把每個源點串連到一點大源點上，容量為無窮。

把每個匯點串連到一點大匯點上，容量為無窮。

求出的最大流=最小割=答案。

#include <iostream>#include<cstdio>#include<cstring>#include<cstdio>#include<queue>#include<vector>using namespace std;const int maxint=1000001;#define maxn 40008int s,t;struct edgee{int from,to,cap,flow;};vector<edgee> edges;vector<int> G[maxn];bool vis[maxn];int dist[maxn];int cur[maxn];void add(int from,int to,int cap){edges.push_back((edgee){from,to,cap,0});edges.push_back((edgee){to,from,cap,0});int k=edges.size();G[from].push_back(k-2);//奇數為正向弧 G[to].push_back(k-1);}int max(int a,int b){return a>b?a:b;}int min(int a,int b){return a<b?a:b;}bool bfs(){memset(vis,0,sizeof(vis));queue<int> q;dist[s]=0;vis[s]=1;q.push(s);while(!q.empty()){int u=q.front();q.pop();int sz=G[u].size();for(int i=0;i<sz;i++){edgee& e=edges[G[u][i]];if(!vis[e.to] && e.cap>e.flow){vis[e.to]=1;dist[e.to]=dist[u]+1;q.push(e.to);}}}return vis[t];}int dfs(int u,int low){if(u==t || low==0)return low;int flow=0,sz=G[u].size(),d;//&引用的妙用for(int& i=cur[u];i<sz;i++) //正在考慮的最後一條邊開始考慮（效率關鍵）{edgee& e=edges[G[u][i]];if( dist[u]+1==dist[e.to] && (d=dfs(e.to,min(e.cap-e.flow,low)))>0 ){e.flow+=d;edges[G[u][i]^1].flow-=d;flow+=d;low-=d;if(low==0)break;}}return flow;}int Dinic(){int flow=0;while(bfs()){memset(cur,0,sizeof(cur));flow+=dfs(s,maxint);}return flow;}int main(){int i,j;int flag=1;int n,m;while(~scanf("%d%d",&n,&m)){edges.clear();for(i=0;i<maxn;i++)G[i].clear();s=40002,t=40003;for(i=0;i<n;i++){int k;for(j=0;j<m;j++){scanf("%d",&k);if(k==0){if(j!=m-1)add(i*m+j,i*m+j+1,1);}else{if(k==2){if(j!=m-1){add(i*m+j,i*m+j+1,1);}add(s,i*m+j,10001);}elseif(k==1){if(j!=m-1){add(i*m+j,i*m+j+1,1);}add(i*m+j,t,10001);}}}}for(j=0;j<m;j++)for(i=0;i<n-1;i++){add((i)*m+j,(i+1)*m+j,1);}int temp=Dinic();    printf("Case %d:\n%d\n",flag++,temp);}return 0;}