HDU 5988.Coding Contest 最小费用最大流
Coding Contest
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1751 Accepted Submission(s): 374
For the i-th path, the wires have been stabilized at first and the first competitor who walker through it would not break the wires. Since then, however, when a person go through the i - th path, there is a chance of pi to touch
the wires and affect the whole networks. Moreover, to protect these wires, no more than ci competitors are allowed to walk through the i-th path.
Now you need to find a way for all competitors to get their lunch, and minimize the possibility of network crashing.
For each test case, the first line consists of two integers N (N ≤ 100) and M (M ≤ 5000). Each of the next N lines contains two integers si and bi (si , bi ≤ 200).
Each of the next M lines contains three integers ui , vi and ci(ci ≤ 100) and a float-point number pi(0 < pi < 1).
It is guaranteed that there is at least one way to let every competitor has lunch.
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
using namespace std;
typedef long long ll;
typedef pair<double,int> P;
#define PI acos(-1.0)
const int maxn=,maxm=1e5+,inf=0x3f3f3f3f,mod=1e9+;
const ll INF=1e18+;
const double eps=0.000001;
struct edge
{
int from,to;
int cap;
double cost;
int rev;
};
int NN;
vector<edge>G[maxn];
double h[maxn];
///顶点的势,取h(u)=(s到u的最短距离),边e=(u,v)的长度变成d`(e)=d(e)+h(u)-h(v)>=0
double dist[maxn];
int prevv[maxn],preve[maxn];///前驱结点和对应的边
void addedge(int u,int v,int cap,double cost)
{
edge e;
e.from=u,e.to=v,e.cap=cap,e.cost=cost,e.rev=G[v].size();
G[u].push_back(e);
e.from=v,e.to=u,e.cap=,e.cost=-cost,e.rev=G[u].size()-;
G[v].push_back(e);
}
double min_cost_flow(int s,int t,int f)
{
double res=0.0;
fill(h,h+NN,0.0);
while(f>)
{
priority_queue<P,vector<P>,greater<P> >q;
fill(dist,dist+NN,inf);
dist[s]=0.0;
q.push(P(dist[s],s));
while(!q.empty())
{
P p=q.top();
q.pop();
int u=p.second;
if(dist[u]<p.first) continue;
for(int i=; i<G[u].size(); i++)
{
edge e=G[u][i];
if(e.cap>&&dist[e.to]-(dist[u]+e.cost+h[u]-h[e.to])>=eps)
{
dist[e.to]=dist[u]+e.cost+h[u]-h[e.to];
prevv[e.to]=u;
preve[e.to]=i;
q.push(P(dist[e.to],e.to));
}
}
}
if(fabs(dist[t]-inf)<=eps) return res;
for(int i=; i<NN; i++) h[i]+=dist[i];
int d=f;
for(int i=t; i!=s; i=prevv[i])
d=min(d,G[prevv[i]][preve[i]].cap);
f-=d;
res+=d*h[t];
//cout<<d<<" "<<h[t]<<" "<<d*h[t]<<endl;
for(int i=t; i!=s; i=prevv[i])
{
//cout<<i<<" ";
edge &e=G[prevv[i]][preve[i]];
e.cap-=d;
G[i][e.rev].cap+=d;
}
//cout<<s<<endl;
}
return res;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n,m;
scanf("%d%d",&n,&m);
int s=,t=n+,f;
NN=t+;
for(int i=; i<=n; i++)
{
int a,b;
scanf("%d%d",&a,&b);
int x=min(a,b);
a-=x,b-=x;
if(a>) addedge(s,i,a,);
else if(b>) addedge(i,t,b,);
}
for(int i=; i<=m; i++)
{
int u,v,cap;
double cost;
scanf("%d%d%d%lf",&u,&v,&cap,&cost);
if(cap>) addedge(u,v,,);
if(cap->) addedge(u,v,cap-,-*log(1.0-cost));
}
double sum=min_cost_flow(s,t,inf);
printf("%.2f\n",1.0-exp(-1.0*sum));
for(int i=; i<NN; i++) G[i].clear();
}
return ;
}
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