uva 10330 - Power Transmission(网络流）

uva 10330 - Power Transmission

题目大意：最大流问题。

解题思路：増广路算法。

#include <stdio.h>

#include <string.h>

#include <queue>

using namespace std;

#define min(a,b) (a)<(b)?(a):(b)

const int N = 105;

const int INF = 0x3f3f3f3f;

int n, s[N], g[N][N], f[N][N];

void init() {

	memset(s, 0, sizeof(s));

	memset(g, INF, sizeof(g));

	memset(f, INF, sizeof(f));

	for (int i = 1; i <= n; i++)

		scanf("%d", &s[i]);

	int a, b, c, t;

	scanf("%d", &t);

	for (int i = 0; i < t; i++) {

		scanf("%d%d%d", &a, &b, &c);

		g[a][b] = min(g[a][b], c);

		f[a][b] = 0;

	}

	scanf("%d%d", &a, &b);

	for (int i = 0; i < a; i++) {

		scanf("%d", &c);

		g[0][c]--;

		f[0][c] = 0;

	}

	for (int i = 0; i < b; i++) {

		scanf("%d", &c);

		g[c][n + 1]--;

		f[c][n + 1] = 0;

	}

}

int solve() {

	n++;

	s[n] = INF;

	int a[N], vis[N];

	int ans = 0;

	queue<int> q;

	while (1) {

		memset(vis, 0, sizeof(vis));

		memset(a, 0, sizeof(a));

		int c = 0, t;

		vis[c] = 0;

		a[c] = INF;

		q.push(c);

		while (!q.empty()) {

			c = q.front(), q.pop();

			for (int i = 1; i <= n; i++) {

				if (g[c][i] == INF) continue;

				t = min(g[c][i] - f[c][i], min(a[c], s[i]));

				if (t > a[i]) {

					vis[i] = c;

					a[i] = t;

					q.push(i);

				}

			}

		}

		if (a[n] == 0) break;

		ans += a[n];

		for (int i = n; i; i = vis[i]) {

			s[i] -= a[n];

			f[vis[i]][i] += a[n];

		}

	}

	return ans;

}

int main () {

	while (scanf("%d", &n) == 1) {

		init();

		printf("%d\n", solve());

	}

	return 0;

}