题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5988
题目思路:最小费用最大流模板题,只不过有很多细节要注意,首先,对于每条边的第一次费用是0,将一条边非为1和cap-1,分别添加进入,第二,费用要用-log(1.pi)就是最小费用了,第三,要设置首尾,第四,必须要用eps进行判断,否则会超时
AC代码
#include<iostream>
#include<cmath>
#include<queue>
#include<cstring>
using namespace std;
const double eps=1e-8;
#define maxn 10007
#define maxm 5007
int first[100000];
int edge_num;
struct edge{
int flow;
int cap;
int u,v;
int next;
double cost;
}e[100000];
double dis[maxn];
bool vis[maxn];
int p[maxn];
int n,m;
double ans;
void add_edge(int u,int v,double w ,int c)
{
e[edge_num].u=u;
e[edge_num].v=v;
e[edge_num].cost=w;
e[edge_num].cap=c;
e[edge_num].flow=0;
e[edge_num].next=first[u];
first[u]=edge_num++;
e[edge_num].u=v;
e[edge_num].v=u;
e[edge_num].cost=-w;
e[edge_num].cap=0;
e[edge_num].flow=0;
e[edge_num].next=first[v];
first[v]=edge_num++;
}
void EK(int s,int t)
{
while(1)
{
queue<int>q;
q.push(s);
for(int i=0;i<n+30;++i)
{
dis[i]=1000000000007;
vis[i]=false;
p[i]=-1;
}
dis[s]=0;
while(!q.empty())
{
int u=q.front();
q.pop();
vis[u]=false;
for(int k=first[u];k!=-1;k=e[k].next)
{
int v=e[k].v;
if(e[k].cap>e[k].flow&&dis[v]-dis[u]-e[k].cost>eps)
{
dis[v]=dis[u]+e[k].cost;
p[v]=k;
if(!vis[v])
{
vis[v]=true;
q.push(v);
}
}
}
}
if(p[t]==-1)
break;
int a=999999999;
for(int k=p[t];k!=-1;k=p[e[k].u])
a=min(a,e[k].cap-e[k].flow);
for(int k=p[t];k!=-1;k=p[e[k].u])
{
e[k].flow+=a;
e[k^1].flow-=a;
}
ans+=dis[t]*a;
}
}
int main()
{
int t;
scanf("%d",&t);
int temp[maxn];
while(t--)
{
memset(first,-1,sizeof(first));
ans=0;
edge_num=0;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;++i)
{
int x,y;
scanf("%d%d",&x,&y);
temp[i]=x-y;
}
for(int i=1;i<=m;++i)
{
int u,v,c;
double pi;
scanf("%d%d%d%lf",&u,&v,&c,&pi);
if(c>0)
add_edge(u,v,0,1);
if(c-1>0)
add_edge(u,v,-log(1.0-pi),c-1);
}
for(int i=1;i<=n;++i)
{
if(temp[i]>0)
add_edge(0,i,0,temp[i]);
if(temp[i]<0)
add_edge(i,n+1,0,-temp[i]);
}
EK(0,n+1);
printf("%.2f\n",1-1/exp(ans));
}
}