POJ2455 Secret Milking Machine（二分上線構圖求最大流）

一張無向圖中有 N 個點，M 條邊，每條邊都有一個權值，且每條邊只能用一次，要求找出 T 條從 1 到 N 的路徑，使這 T 條路徑所經過的邊中，權值的最大值最小。

思想：二分邊權值上線，多次構圖求最大流，判斷max_flow是否滿足 T 。詳見代碼：

#include<cstdio>#include<cstring>#include<queue>#define find_min(a,b) a<b?a:b#define find_max(a,b) a>b?a:busing namespace std;const int N = 210;const int MAX = (int)10e7;struct Edge{int s,e,v,next;}edge[2*N*N],mat[N*N];int e_num,head[N],d[N],sp,tp;int n,m;void AddEdge(int a,int b,int c){edge[e_num].s=a; edge[e_num].e=b; edge[e_num].v=c;edge[e_num].next=head[a]; head[a]=e_num++;edge[e_num].s=b; edge[e_num].e=a; edge[e_num].v=c;edge[e_num].next=head[b]; head[b]=e_num++;}int bfs(){queue <int> q;memset(d,-1,sizeof(d));d[sp]=0;q.push(sp);while(!q.empty()){int cur=q.front();q.pop();for(int i=head[cur];i!=-1;i=edge[i].next){int u=edge[i].e;if(d[u]==-1 && edge[i].v>0){d[u]=d[cur]+1;q.push(u);}}}return d[tp] != -1;}int dfs(int a,int b){int r=0;if(a==tp)return b;for(int i=head[a];i!=-1 && r<b;i=edge[i].next){int u=edge[i].e;if(edge[i].v>0 && d[u]==d[a]+1){int x=find_min(edge[i].v,b-r);x=dfs(u,x);r+=x;edge[i].v-=x;edge[i^1].v+=x;}}if(!r)d[a]=-2;return r;}int dinic(int sp,int tp){int total=0,t;while(bfs()){while(t=dfs(sp,MAX))total+=t;}return total;}void init(int index){    int i,j;    e_num=0;    memset(head,-1,sizeof(head));    for(i=1;i<=m;i++){        if(mat[i].v<=index)AddEdge(mat[i].s,mat[i].e,1);    }}int main(){    int t_num,i,j,a,b,c;    while(~scanf("%d%d%d",&n,&m,&t_num)){        sp=1;tp=n;        int high=-1,low=MAX;        for(i=1;i<=m;i++){            scanf("%d%d%d",&a,&b,&c);            high=find_max(high,c);            low=find_min(low,c);            mat[i].s=a; mat[i].e=b; mat[i].v=c;        }        int mid,ans;        while(low<=high){            mid=(low+high)/2;            init(mid);            ans=dinic(sp,tp);            if(ans<t_num)low=mid+1;            else high=mid-1;        }        printf("%d\n",low);    }    return 0;}