isap测模板--HDU3549

Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 5990    Accepted Submission(s): 2790

Problem Description

Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for the weighted directed graph.

Input

The first line of input contains an integer T, denoting the number of test cases.
For each test case, the first line contains two integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <= 15, 0 <= M <= 1000)
Next M lines, each line contains three integers X, Y and C, there is an edge from X to Y and the capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)

Output

For each test cases, you should output the maximum flow from source 1 to sink N.

Sample Input

2
3 2
1 2 1
2 3 1
3 3
1 2 1
2 3 1
1 3 1

Sample Output

Case 1: 1
Case 2: 2

/***************************************** author:crazy_石头* algorithm:sap+当前弧+gap优化* date:2013/09/29*****************************************/#include<cstdio>#include<cstring>#include<climits>using namespace std;#define N 90005#define M 4000005#define INF INT_MAXint s,t,num,pre[N],dis[N],gap[N],head[N],cur[N];struct node{     int u,v,w;     int next;}e[M];inline void add(int u,int v,int w){     e[num].u=u;     e[num].v=v;     e[num].w=w;     e[num].next=head[u];     head[u]=num++;}inline void addedge(int u,int v,int w){     add(u,v,w);     add(v,u,0);}inline int max(int a,int b){    return a>b?a:b;}inline int min(int a,int b){    return a<b?a:b;}int isap(int s,int t,int n){     int top,mindis,maxflow=0,v,i,aug;     bool flag;     for(i=0;i<=n;i++)     {         cur[i]=head[i];         gap[i]=dis[i]=0;     }     top=pre[s]=s;     aug=INF;     while(dis[s]<n)     {         flag=0;         for(i=cur[top];i!=-1;i=e[i].next)         {             v=e[i].v;             if(e[i].w>0&&dis[top]==dis[v]+1)             {                 flag=1;                break;             }         }         if(flag)         {             pre[v]=top;             cur[top]=i;             aug=min(aug,e[i].w);             top=v;             if(top==t)             {                 while(top!=s)                 {                     top=pre[top];                     e[cur[top]].w-=aug;                     e[cur[top]^1].w+=aug;                 }                 top=s;                 maxflow+=aug;                 aug=INF;             }         }         else         {             if(--gap[dis[top]]==0)                break;              mindis=n;             cur[top]=head[top];             for(i=head[top];i!=-1;i=e[i].next)             {                 v=e[i].v;                 if(e[i].w>0&&dis[v]+1<mindis)                 {                     mindis=dis[v]+1;                     cur[top]=i;                 }             }             dis[top]=mindis;             gap[mindis]++;             if(top!=s)                 top=pre[top];         }     }     return maxflow;}void init(){     num=0;     memset(head,-1,sizeof(head));}int main(){     int nodenum,m,n,test;     scanf("%d",&test);     for(int ii=1;ii<=test;ii++)     {         scanf("%d%d",&n,&m);         init();         while(m--)         {             int u,v,w;             scanf("%d%d%d",&u,&v,&w);             addedge(u,v,w);         }        //source=1,sink=n,nodenum=n;         int res=isap(1,n,n);         printf("Case %d: %d\n",ii,res);     }     return 0;}