HDU 3549 最大流 水题

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

23 21 2 12 3 13 31 2 12 3 11 3 1

 

Sample Output

Case 1: 1Case 2: 2

 

Author

HyperHexagon

 

Source

HyperHexagon's Summer Gift (Original tasks)

 

/************************************************Desiner:hltime:2015/11/02Exe.Time:639MSExe.Memory:3184K题解：裸最大流。做做练手速 不过还是小心。我一开始用了sap(S,V,V+1)导致答案出不来。 ************************************************/#include <iostream>    #include <algorithm>    #include <cstring>    #include <string>    #include <cstdio>    #include <cmath>          using namespace std;    const int MAXN = 100010 ; //点数最大值      const int MAXM = 400010 ; //边数最大值      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++;      }        //最大流开始     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;      }      //最大流结束       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;  }