BZOJ3130 SDOI2013 费用流 二分法+网络流

#include <queue>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <climits>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 1000000.0

const int MAXN=100+2;
const int MAXM=1000+2;
struct HASH{
    int u;
    HASH *next;
    HASH(){}
    HASH(int _u,HASH *_next):u(_u),next(_next){}
}*tab[MAXN],*cur[MAXN],mem[2*MAXM];
struct EDGE{
    int u,v;
    double c;
    EDGE(){}
    EDGE(int _u,double _c):u(_u),c(_c){}
}e[2*MAXM],t[MAXM];
bool operator<(EDGE a,EDGE b){ return a.c<b.c;}
int N,M,P,A,L=INT_MAX,U=INT_MIN,cnt,d[MAXN];
queue<int> q;

void Insert(int u,int v,double c){
    tab[u]=&(mem[cnt]=HASH(cnt,tab[u])),e[cnt++]=EDGE(v,c);
    tab[v]=&(mem[cnt]=HASH(cnt,tab[v])),e[cnt++]=EDGE(u,0);
}

bool BFS(int S,int T){
    memset(d,-1,sizeof(d));
    d[S]=0,q.push(S);

    int x;
    while(!q.empty()){
        x=q.front(),q.pop();
        for(HASH *p=tab[x];p;p=p->next)
            if(e[p->u].c && d[e[p->u].u]<0)
                d[e[p->u].u]=d[x]+1,q.push(e[p->u].u);
    }
    return d[T]>0;
}

double DFS(int x,double f,int t){
    if(x==t) return f;

    double used=0,flow;
    for(HASH *p=cur[x];p;p=p->next)
        if(e[p->u].c && d[e[p->u].u]==d[x]+1){
            flow=DFS(e[p->u].u,min(e[p->u].c,f-used),t);
            e[p->u].c-=flow,e[p->u^1].c+=flow,used+=flow;
            if(e[p->u].c) cur[x]=p;
            if(used==f) return f;
        }

    if(!used) d[x]=-1;
    return used;
}

int Dinic(int S,int T){
    double ret=0;
    while(BFS(S,T)){
        memcpy(cur,tab,sizeof(cur));
        ret+=DFS(S,INF,T);
    }
    return ret;
}

bool Check(double x){
    memset(tab,0,sizeof(tab)),cnt=0;
    for(int i=1;i<=M;i++)
        Insert(t[i].u,t[i].v,min(t[i].c,x));
    return Dinic(1,N)==A;
}

int main(){
    scanf("%d %d %d",&N,&M,&P);
    for(int i=1;i<=M;i++){
        scanf("%d %d %lf",&t[i].u,&t[i].v,&t[i].c);
        L=min(L,(int)t[i].c),U=max(U,(int)t[i].c);
        Insert(t[i].u,t[i].v,t[i].c);
    }

    A=Dinic(1,N),sort(t+1,t+M+1);

    double l=0.0,r=U,m;
    while(r-l>1e-6){
        m=(l+r)/2;
        if(Check(m)) r=m;
        else l=m;
    };
    printf("%d\n%.5lf\n",A,(double)P*m);

    return 0;
}