poj2455 Secret Milking Machine（二分答案+最大流）

二分答案，双向边网络流，反向边容量直接设为val即可。可以选择的边容量为1，跑最大流，看是否满流（是否存在K条路径）。

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;#define ll long long#define inf 0x3f3f3f3f#define N 210inline int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();    return x*f;}int n,m,K,h[N],num=1,lev[N],T=201,cur[N];struct edge{    int to,next,val;}data[90000];struct Edge{    int x,y,val;}e[40010];inline bool cmp(Edge x,Edge y){return x.val<y.val;}inline void add(int x,int y,int val){    data[++num].to=y;data[num].next=h[x];h[x]=num;data[num].val=val;    data[++num].to=x;data[num].next=h[y];h[y]=num;data[num].val=val;}inline bool bfs(){    queue<int>q;memset(lev,0,sizeof(lev));    q.push(0);lev[0]=1;    while(!q.empty()){        int x=q.front();q.pop();        for(int i=h[x];i;i=data[i].next){            int y=data[i].to;if(lev[y]||!data[i].val) continue;            lev[y]=lev[x]+1;q.push(y);        }    }return lev[T];}inline int dinic(int x,int low){    if(x==T) return low;int tmp=low;    for(int i=h[x];i;i=data[i].next){        int y=data[i].to;if(lev[y]!=lev[x]+1||!data[i].val) continue;        int res=dinic(y,min(tmp,data[i].val));        if(!res) lev[y]=0;else tmp-=res,data[i].val-=res,data[i^1].val+=res;        if(!tmp) return low;    }return low-tmp;}inline bool jud(int mid){    memset(h,0,sizeof(h));num=1;int ans=0;    add(0,1,K);add(n,T,K);    for(int i=1;i<=m;++i)        if(e[i].val<=mid) add(e[i].x,e[i].y,1);    while(bfs()){memcpy(cur,h,sizeof(cur));ans+=dinic(0,inf);}    return ans==K;}int main(){//  freopen("a.in","r",stdin);    n=read();m=read();K=read();    for(int i=1;i<=m;++i) e[i].x=read(),e[i].y=read(),e[i].val=read();    sort(e+1,e+m+1,cmp);    int l=1,r=1e6;    while(l<=r){        int mid=l+r>>1;        if(jud(mid)) r=mid-1;        else l=mid+1;    }printf("%d\n",r+1);    return 0;}