HDU 3549 Flow Problem(入门) [最大流]

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                                            Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 21871    Accepted Submission(s): 10294

 

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)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>

using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 15 + 10;
const int maxm = 1000 + 10;

int n,m;
int l[maxn];//记录层数
int h[maxn];//链式前向星
int tot = 0;

struct edge
{
  int to;
  int c;
  int next;
  edge(int x = 0, int y = 0, int z = 0) : to(x), c(y), next(z) {}
 }es[maxm*2];//记录边 注意是2倍

void add_edge(int u, int v, int c)
{
    es[tot] = edge(v,c,h[u]);
    h[u] = tot++;
}

bool bfs(int s, int t)
{
   memset(l,0,sizeof(l));
   l[s] = 1;
   queue <int> q;
   q.push(1);
   while(!q.empty())
   {
    int u = q.front();
    q.pop();
    if(u == t)  return true;
    for(int i = h[u]; i != -1; i = es[i].next)
        {
         int v = es[i].to;
         if(!l[v] && es[i].c) {l[v] = l[u] + 1; q.push(v);}
        }
   }
   return false;
}

int dfs(int x, int t, int mf)
{
    if(x == t) return mf;
    int ret = 0;
    for(int i = h[x]; i != -1; i = es[i].next)
    {
      if(es[i].c && l[x] == l[es[i].to] - 1)
      {
        int f = dfs(es[i].to,t,min(es[i].c,mf - ret));
        es[i].c -= f;
        es[i^1].c += f;
        ret += f;
        if(ret == mf) return ret;
      }
    }
    return ret;
}

int dinic(int s, int t)
{
  int ans = 0;
  while(bfs(s,t))  ans += dfs(s,t,INF);
  return ans;
}

int main()
{
  int T;
  scanf("%d",&T);
  int TT = T;
   while(T--)
   {
   scanf("%d%d",&n,&m);
   tot = 0;
   memset(h,-1,sizeof(h));
   int u,v,c;
   for(int i = 0; i < m; i++)
   {
    scanf("%d%d%d",&u,&v,&c);
    add_edge(u,v,c);
    add_edge(v,u,0);//增加反向边
   }
   int ans = dinic(1,n);
   printf("Case %d: %d\n",TT-T,ans);
   }
   return 0;
}