题目：http://acm.hdu.edu.cn/showproblem.php?pid=3549
题意：给定一个有向图，有n个点和m条边，每条边有一个容量，求从1到n的最大流量。输入形式：首先一个t，代表数据组数，然后两个数n，m，接下来m行，形式如a b c，意为从a到b有一条容量为c的边
思路：最大流模板题，我用的dinic算法
#include #include #include #include #include #include #include #include using namespace std; const int N = 30; const int INF = 0x3f3f3f3f; struct edge { int to, cap, next; }g[N*100]; int level[N], iter[N], head[N]; int s[N][N]; int n, m, cnt, _case = 0; void add_edge(int v, int u, int cap) { g[cnt].to = u, g[cnt].cap = cap, g[cnt].next = head[v], head[v] = cnt++; g[cnt].to = v, g[cnt].cap = 0, g[cnt].next = head[u], head[u] = cnt++; } bool bfs(int s, int t) { memset(level, -1, sizeof level); queue que; que.push(s); level[s] = 0; while(! que.empty()) { int v = que.front(); que.pop(); for(int i = head[v]; i != -1; i = g[i].next) { int u = g[i].to; if(g[i].cap > 0 && level[u] < 0) { level[u] = level[v] + 1; que.push(u); } } } return level[t] == -1; } int dfs(int v, int t, int f) { if(v == t) return f; for(int &i = iter[v]; i != -1; i = g[i].next) /*当前弧优化*/ { int u = g[i].to; if(g[i].cap > 0 && level[v] < level[u]) { int d = dfs(u, t, min(f, g[i].cap)); if(d > 0) { g[i].cap -= d, g[i^1].cap += d; return d; } } } return 0; } int dinic(int s, int t) { int flow = 0, f; while(true) { if(bfs(s, t)) return flow; memcpy(iter, head, sizeof head); while(f = dfs(s, t, INF), f > 0) flow += f; } } int main() { int t, a, b, c; scanf("%d", &t); while(t--) { scanf("%d%d", &n, &m); memset(head, -1, sizeof head); cnt = 0; for(int i = 0; i < m; i++) { scanf("%d%d%d", &a, &b, &c); add_edge(a, b, c); } printf("Case %d: %d\n", ++_case, dinic(1, n)); } return 0; }
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