HDU 3549 最大流 水題

Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 11558    Accepted Submission(s): 5471

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

HyperHexagon

 

Source

HyperHexagon's Summer Gift (Original tasks)

 

/************************************************

Desiner:hl
time:2015/11/02
Exe.Time:639MS
Exe.Memory:3184K



題解：裸最大流。做做練手速 
不過還是小心。我一開始用了sap(S,V,V+1)導致答案出不來。 

************************************************/

#include <iostream>    
#include <algorithm>    
#include <cstring>    
#include <string>    
#include <cstdio>    
#include <cmath>      
    
using namespace std;    
const int MAXN = 100010 ; //點數最大值      
const int MAXM = 400010 ; //邊數最大值      
const int INF = 0x3f3f3f3f;      
  
int S,V,N,M;  

 
struct Edge{      
    int to,next,cap,flow;      
}edge[MAXM];//注意是MAXM      
int tol;      
int head[MAXN];      
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];       

    
void init(){      
    tol = 0;      
    memset(head,-1,sizeof(head));      
}       
    
void addedge(int u,int v,int w,int rw=0){      
    edge[tol].to = v;      
    edge[tol].cap = w;      
    edge[tol].next = head[u];      
    edge[tol].flow = 0;      
    head[u] = tol++;      
    edge[tol].to = u;      
    edge[tol].cap = rw;      
    edge[tol].next = head[v];      
    edge[tol].flow = 0;      
    head[v] = tol++;      
}      
  //最大流開始     
int sap(int start,int end,int N){      
    memset(gap,0,sizeof(gap));      
    memset(dep,0,sizeof(dep));      
    memcpy(cur,head,sizeof(head));      
    int u = start;      
    pre[u] = -1;      
    gap[0] = N;      
    int ans = 0;      
    while(dep[start] < N){      
        if(u==end){      
            int Min = INF;      
            for(int i=pre[u];i!= -1; i=pre[edge[i^1].to])      
                if(Min > edge[i].cap - edge[i].flow)      
                    Min = edge[i].cap - edge[i].flow;      
                          
            for(int i=pre[u];i!=-1;i=pre[edge[i^1].to]){      
                edge[i].flow += Min;      
                edge[i^1].flow -=Min;      
            }      
            u=start;      
            ans +=Min;      
            continue;      
        }      
        bool flag = false;      
        int v;      
        for(int i= cur[u];i!=-1;i=edge[i].next){      
            v=edge[i].to;      
            if(edge[i].cap-edge[i].flow&&dep[v]+1==dep[u]){      
                flag=true;      
                cur[u]=pre[v]=i;      
                break;      
            }      
        }      
        if(flag){      
            u=v;      
            continue;      
        }      
        int Min = N;      
        for(int i=head[u];i!= -1;i=edge[i].next)      
            if(edge[i].cap-edge[i].flow&&dep[edge[i].to]<Min){      
                Min=dep[edge[i].to];      
                cur[u] = i;      
            }      
        gap[dep[u]]--;      
        if(!gap[dep[u]]) return ans;      
        dep[u] = Min +1;      
        gap[dep[u]]++;      
        if(u!=start) u = edge[pre[u]^1].to;      
    }      
    return ans;      
}      
//最大流結束     
  
int build(){  
    int i,j,k,l,m,n,st,en,va;  
    init();  
    for(i=1;i<=M;i++){  
    	scanf("%d%d%d",&st,&en,&va);
    	addedge(st,en,va);
    } 

    int orz=sap(S,V,V);  
    return orz;  
}  
int main(){     

	int T;
    int m,n,q,p;      
    int i,j,k,a,b,c; 
	int pisum;  
	scanf("%d",&T);
    for(int cas=1;cas<=T;cas++){
    	scanf("%d%d",&N,&M);  
		S=1;
		V=N;		
    	printf("Case %d: %d\n",cas,build());
    }	 
    return 0;  
}