Coding Contest
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1751 Accepted Submission(s): 374
Problem Description
A coding contest will be held in this university, in a huge playground. The whole playground would be divided into N blocks, and there would be M directed paths linking these blocks. The i-th path goes from the ui-th block to the vi-th block. Your task is to solve the lunch issue. According to the arrangement, there are sicompetitors in the i-th block. Limited to the size of table, bi bags of lunch including breads, sausages and milk would be put in the i-th block. As a result, some competitors need to move to another block to access lunch. However, the playground is temporary, as a result there would be so many wires on the path.
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 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.
Input
The first line of input contains an integer t which is the number of test cases. Then t test cases follow.
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.
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.
Output
For each turn of each case, output the minimum possibility that the networks would break down. Round it to 2 digits.
Sample Input
1
4 4
2 0
0 3
3 0
0 3
1 2 5 0.5
3 2 5 0.5
1 4 5 0.5
3 4 5 0.5
Sample Output
0.50
Source
题意：有n个区域，每个区域有一些人数si和食物bi，区域之间有m条定向路径，每条路径有人数通过上限ci。路径之间铺了电线，每当有人通过路径时有pi的概率会触碰到电线，但是第一个通过的人一定不会触碰到电线。求每个人都通过路径获取到食物后触碰到电线的最小概率。
思路：首先每个区域的一些人都直接获取自己区域的食物，如果人过量，源点链接本区域，容量是过量的人数，费用是0；如果食物过量，本区域连接汇点，容量是过量的食物，费用是0。剩下的人通过路径到其他区域获取食物，m条路径之间建立边，首先是一定不会触碰，容量是1，花费为0,；接下来的才是容量为ci-1的边，并且还拥有一个费用，这就是一个最小费用最大流的问题，但是这个费用怎么转换呢？把每个概率取log(1-pi)，并且取负数-log(1-pi)。这样的话最小花费(-log(x1))+(-log(x2))+(-log(x3))+....=-(log(x1)+log(x2)+log(x3)+...)=-log(x1*x2*x3...)。最后最小费用ans取负数再作为e的指数，即exp(-ans)是最大乘积。
代码：
#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 ;
}
最小费用最大流