Coding Contest(费用流变形题，double)

http://acm.hdu.edu.cn/showproblem.php?pid=5988

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 5337 Accepted Submission(s): 1256

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

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

< 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

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

2016ACM/ICPC亚洲区青岛站-重现赛（感谢中国石油大学）

求网络被破坏的最小可能性，因为是乘法，所以要用取对数的方法把它改成加法。

因为概率是小于1的，所以取对数完是负数，需要用 - 把它转为正数。

但是转为正数后，原来的最小值就会变为最大值，所以用p=-log2(1-p)，转为求不被破坏的最大可能行，跑费用流

因为是浮点型，所以在松弛的时候要加上eps。

最后，要用for去代替memset,不然可能会t...

  1 #include<iostream>
  2 #include<algorithm>
  3 #include<queue>
  4 #include<cstring>
  5 #include<cmath>
  6 #include<cstdio>
  7 using namespace std;
  8 
  9 const double eps=1e-9;
 10 const int INF=0x3f3f3f3f;
 11 const int N=500005;
 12 const int M=500005;
 13 int top;
 14 double dist[N];
 15 int pre[N];
 16 bool vis[N];
 17 int c[N];
 18 int maxflow;
 19 
 20 struct Vertex{
 21     int first;
 22 }V[N];
 23 struct Edge{
 24     int v,next;
 25     int cap,flow;
 26     double cost;
 27 }E[M];
 28 
 29 void init(int num){
 30    // memset(V,-1,sizeof(V));
 31     for(int i=0;i<num;i++){
 32         V[i].first=-1;
 33     }
 34     top=0;
 35     maxflow=0;
 36 }
 37 
 38 void add_edge(int u,int v,int c,double cost){
 39     E[top].v=v;
 40     E[top].cap=c;
 41     E[top].flow=0;
 42     E[top].cost=cost;
 43     E[top].next=V[u].first;
 44     V[u].first=top++;
 45 }
 46 
 47 void add(int u,int v,int c,double cost){
 48     add_edge(u,v,c,cost);
 49     add_edge(v,u,0,-cost);
 50 }
 51 
 52 bool SPFA(int s,int t,int n){
 53     int i,u,v;
 54     queue<int>qu;
 55    // memset(vis,false,sizeof(vis));
 56   //  memset(c,0,sizeof(c));
 57    // memset(pre,-1,sizeof(pre));
 58     for(i=0;i<=n+2;i++){
 59         dist[i]=INF;
 60         vis[i]=false;
 61         c[i]=0;
 62         pre[i]=-1;
 63     }
 64   //  memset(dist,INF,sizeof(dist));
 65     vis[s]=true;
 66     c[s]++;
 67     dist[s]=0;
 68     qu.push(s);
 69     while(!qu.empty()){
 70         u=qu.front();
 71         qu.pop();
 72         vis[u]=false;
 73         for(i=V[u].first;~i;i=E[i].next){
 74             v=E[i].v;
 75             if(E[i].cap>E[i].flow&&dist[v]>dist[u]+E[i].cost+eps){
 76                 dist[v]=dist[u]+E[i].cost;
 77                 pre[v]=i;
 78                 if(!vis[v]){
 79                     c[v]++;
 80                     qu.push(v);
 81                     vis[v]=true;
 82                     if(c[v]>n){
 83                         return false;
 84                     }
 85                 }
 86             }
 87         }
 88     }
 89     if(dist[t]==INF){
 90         return false;
 91     }
 92     return true;
 93 }
 94 
 95 double MCMF(int s,int t,int n){
 96     int d,i;
 97     double mincost=0;
 98     while(SPFA(s,t,n)){
 99         d=INF;
100         for(i=pre[t];~i;i=pre[E[i^1].v]){
101             d=min(d,E[i].cap-E[i].flow);
102         }
103         maxflow+=d;
104         for(i=pre[t];~i;i=pre[E[i^1].v]){
105             E[i].flow+=d;
106             E[i^1].flow-=d;
107         }
108         mincost+=dist[t]*d;
109     }
110     return mincost;
111 }
112 
113 int main(){
114     int n,m;
115     int T;
116     scanf("%d",&T);
117     while(T--){
118         scanf("%d %d",&n,&m);
119         init(n+3);
120         int a,b,c;
121         double p;
122         int s=0,t=n+1;
123         for(int i=1;i<=n;i++){
124             scanf("%d %d",&a,&b);
125             if(a>b){
126                 add(s,i,a-b,0);
127             }
128             else if(a<b){
129                 add(i,t,b-a,0);
130             }
131         }
132         for(int i=1;i<=m;i++){
133             scanf("%d %d %d %lf",&a,&b,&c,&p);
134             if(c>0) add(a,b,1,0);
135             if(c>1) add(a,b,c-1,-log2(1-p));
136         }
137         double ans=MCMF(s,t,n+2);
138         printf("%.2f\n",1.0-pow(2,-ans));
139     }
140 }

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