思路:
求一遍网络流 在残余网络上DFS 从起点DFS 从终点把边反向DFS 一个边跟起点连通 跟终点反向的边连通 ans++注:此题不能用tarjan 因为有边权为0的边
//By SiriusRen#include#include #include #include using namespace std;#define N 555#define M 10050int n,m,xx[M],yy[M],zz[M],first[N],vis[N],next[M],v[M],w[M],tot=2;int dfn[N],low[N],cnt,stk[N],top,T,jy,p[N],ans;void Add(int x,int y,int z){w[tot]=z,v[tot]=y,next[tot]=first[x],first[x]=tot++;}void add(int x,int y,int z){Add(x,y,z),Add(y,x,0);}bool tell(){ memset(vis,-1,sizeof(vis)),vis[0]=0; queue q;q.push(0); while(!q.empty()){ int t=q.front();q.pop(); for(int i=first[t];~i;i=next[i]) if(w[i]&&vis[v[i]]==-1) q.push(v[i]),vis[v[i]]=vis[t]+1; }return vis[n]!=-1;}int zeng(int x,int y){ if(x==n)return y; int r=0; for(int i=first[x];~i&&y>r;i=next[i]) if(vis[v[i]]==vis[x]+1&&w[i]){ int t=zeng(v[i],min(y-r,w[i])); w[i]-=t,w[i^1]+=t,r+=t; } if(!r)vis[x]=-1; return r;}void dfs(int x,bool f,int col){ vis[x]=col; for(int i=first[x];~i;i=next[i]) if(!vis[v[i]]&&w[i^f]) dfs(v[i],f,col);}int main(){ memset(first,-1,sizeof(first)); scanf("%d%d",&n,&m),n--; for(int i=1;i<=m;i++)scanf("%d%d%d",&xx[i],&yy[i],&zz[i]),add(xx[i],yy[i],zz[i]); while(tell())while(zeng(0,0x3fffffff)); memset(vis,0,sizeof(vis)),dfs(0,0,1),dfs(n,1,2); for(int i=1;i<=m;i++)if(vis[xx[i]]==1&&vis[yy[i]]==2&&!w[i*2])ans++; printf("%d\n",ans);}