bzoj 1733: [Usaco2005 feb]Secret Milking Machine 神秘的挤奶机 二分+最大流

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二分答案，每次把边长小于mid的边放到图中，正反的边容量都为1，跑最大流

然后用最大流和T判断一下，如果大于T返回true，否则false

代码：

#include<iostream>#include<cstdio>#include<vector>#include<queue>#include<cstring> #define ll long long#define inf 10000001 using namespace std; struct node{    int to,num;}edge[80080];int fro[80080],to[80080],len[80080];int n,m,T;int pn=1;vector<int>v[220];int edgenum=0;int d[220]; void link(int x,int y,int z){    node t;    t.to=y;    t.num=z;    edge[edgenum]=t;    v[x].push_back(edgenum);    edgenum++;    t.to=x;    t.num=z;    edge[edgenum]=t;    v[y].push_back(edgenum);    edgenum++;}  bool bfs(){    memset(d,0,sizeof(d));    d[1]=1;    queue<int>q;    q.push(1);    while(!q.empty()){        int t=q.front();        q.pop();        if(t==pn)continue;        for(int i=0; i<v[t].size(); i++){            int x=v[t][i];            int to=edge[x].to;            int num=edge[x].num;            if(num==0 || d[to]!=0)continue;            d[to]=d[t]+1;            q.push(to);        }    }    if(d[pn]!=0)return true;    return false;} ll dfs(int x,int y){    if(x==pn || y==0)return y;    ll sum=0;    for(int i=0; i<v[x].size(); i++){        int j=v[x][i];        int to=edge[j].to;        int num=edge[j].num;        if(num){            if(d[to]!=d[x]+1)continue;            ll k=dfs(to,min(num,y));            y-=k;            sum+=k;            edge[j].num-=k;            edge[j^1].num+=k;            if(y==0)break;        }    }    if(sum==0)d[x]=0;    return sum;} bool check(int x){long long ans=0;edgenum=0;for(int i=1; i<=n; i++)v[i].clear();for(int i=1; i<=m; i++){if(len[i]<=x){link(fro[i],to[i],1);}}while(bfs())ans+=dfs(1,inf);return ans>=T;}int main(){    scanf("%d%d%d",&n,&m,&T);    for(int i=1; i<=m; i++){    scanf("%d%d%d",&fro[i],&to[i],&len[i]);    }    pn=n;    int l=1,r=1000000;    int ans;    while(l<=r){    int mid=(l+r)/2;    if(check(mid))ans=mid,r=mid-1;    else l=mid+1;    }    printf("%d\n",ans);         return 0;}