题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3549

思路分析:该问题为裸的最大网络流问题,数据量不大,使用EdmondsKarp算法求解即可;需要注意的是该问题的点最多有15个,边的数目最多有1000个,所以该图中存在重边,需要将多个重边合为一条边;

代码如下:

#include <queue>
#include <vector>
#include <cstdio>
#include <climits>
#include <cstring>
#include <iostream>
using namespace std; const int MAX_N = ;
int cap[MAX_N][MAX_N], flow[MAX_N][MAX_N];
int a[MAX_N], p[MAX_N]; inline int Min(int a, int b) { return a < b ? a : b; }
int EdmondsKarp(int ver_num)
{
queue<int> q;
int max_flow = ; memset(flow, , sizeof(flow));
for (;;)
{
memset(a, , sizeof(a));
a[] = INT_MAX;
q.push();
while (!q.empty())
{
int u = q.front();
q.pop();
for (int v = ; v <= ver_num; ++v)
{
if (!a[v] && cap[u][v] > flow[u][v])
{
p[v] = u;
q.push(v);
a[v] = Min(a[u], cap[u][v] - flow[u][v]);
}
}
}
if (a[ver_num] == ) break;
for (int u = ver_num; u != ; u = p[u])
{
flow[p[u]][u] += a[ver_num];
flow[u][p[u]] -= a[ver_num];
}
max_flow += a[ver_num];
}
return max_flow;
} int main()
{
int case_times, case_id = ;
int road_num, ver_num; scanf("%d", &case_times);
while (case_times--)
{
int ver_1, ver_2, capa; scanf("%d %d", &ver_num, &road_num);
memset(cap, , sizeof(cap));
for (int i = ; i < road_num; ++i)
{
scanf("%d %d %d", &ver_1, &ver_2, &capa);
cap[ver_1][ver_2] += capa;
}
int ans = EdmondsKarp(ver_num);
printf("Case %d: %d\n", ++case_id, ans);
}
return ;
}

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