【BZOJ3130】费用流

转自老blog

首先说一下 这个题其实就是最大流 只有最大流才可能是答案

费用一定会加在最大流上最大边的 因此二分

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<cstring>
#include<stack>
using namespace std;
double eps=1e-6;
const int INF = 0x7f7f7f7f;
struct edge
{
    int from,to,pre;double cap,flow;
}Edge[2500];
struct edg
{
    int from,to;double cap,flow;
}road[2500];
int cur[200];//当前弧优化
int head[200];
int s,t,n,m,p; 
inline int init()
{
    int now=0;char c;bool flag=false;int ju=1;
    while(1)
    {
        c=getchar();
        if(c=='-')
        {
            ju=-1;
        }
        else if(c>='0'&&c<='9')
        {
            now=now*10+c-'0';
            flag=true;
        }
        else if(flag)return now*ju;
    }
}
int cnt=0;
inline void addedge(int from,int to,double cap)
{
    Edge[++cnt]=((edge){from,to,head[from],cap,0.0});
    head[from]=cnt;
    Edge[++cnt]=((edge){to,from,head[to],0.0,0.0});
    head[to]=cnt;
}
int dis[150];
bool vis[150];
queue<int> q;
bool bfs()
{
    int now;
    while(!q.empty())q.pop();
    memset(vis,false,sizeof(vis));
    vis[s]=true;dis[s]=0;q.push(s);
    while(!q.empty())
    {
        now=q.front();q.pop();
        for(int j=head[now];j;j=Edge[j].pre)
        {
            if(!vis[Edge[j].to]&&Edge[j].cap>Edge[j].flow)
            {
                dis[Edge[j].to]=dis[now]+1;
                vis[Edge[j].to]=true;
                q.push(Edge[j].to);
                if(Edge[j].to==t)return 1;
            }
        }
    }
    return 0;   
}
double dfs(int now,double maxflow)
{
    if(now==t||maxflow==0)return maxflow;
    int &j=cur[now];
    double flow=0.0,f;//可以认为flow是当前节点的总流 f是层流
    for(;j;j=Edge[j].pre)
    {
        if(dis[Edge[j].to]==dis[now]+1&&(f=dfs(Edge[j].to,min(maxflow,Edge[j].cap-Edge[j].flow)))>0)
        {
            flow+=f;
            Edge[j].flow+=f;
            Edge[((j-1)^1)+1].flow-=f;
            maxflow-=f;
            if(maxflow==0.0)break; 
        }
    } 
    return flow;
}
double dinic()
{
    double tot=0.0;
    while(bfs())
    {
        for(int i=s;i<=t;i++)
        {
            cur[i]=head[i];
        }
        tot+=dfs(s,INF*1.0);
    }
    return tot;
}
void make(double mid)
{
    cnt=0;
    memset(head,0,sizeof(head));
    for(int i=1;i<=m;i++)
    {
        addedge(road[i].from,road[i].to,min(road[i].cap,mid));
    }
    return;
}
int main() 
{
    double l=1,mid;
    s=1;
    double r;
    n=init();m=init();p=init();
    t=n;
    int a,b,c;
    for(int i=1;i<=m;i++)
    {
        a=init();b=init();c=init();addedge(a,b,c*1.0);
        road[i].from=a;road[i].to=b;road[i].cap=c;road[i].flow=0.0;
        r=max(r,c*1.0);
    }
    double tmp;
    double ans=dinic();
    while(r-l>eps)
    {
        mid=(l+r)/2;
        make(mid);
        tmp=dinic();
        if(ans-tmp<eps)
        {
            r=mid;
        }
        else l=mid;
    }
    printf("%.0f\n%.4f\n",ans,l*p);
    return 0;
}