POJ 2516 Minimum Cost

题目链接

题意：转一篇题意吧。。感觉写的很详细了，優YoU http://blog.csdn.net/lyy289065406/article/details/6742534

思路：一开始是把所有商家的每种物品和所有供应商所有物品连边跑费用流，结果TLE了，因为这样建出来的图，边数会非常的庞大

那么其实转化一下思路，每种物品之间是不会互相影响的，那么把每种物品求出来后，累加起来就是答案了，然后注意这题数据要读完，直接判断错没读完数据就会WA

代码：

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;

const int MAXNODE = 5005;
const int MAXEDGE = 500005;
typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge {
	int u, v;
	Type cap, flow, cost;
	Edge() {}
	Edge(int u, int v, Type cap, Type flow, Type cost) {
		this->u = u;
		this->v = v;
		this->cap = cap;
		this->flow = flow;
		this->cost = cost;
	}
};

struct MCFC {
	int n, m, s, t;
	Edge edges[MAXEDGE];
	int first[MAXNODE];
	int next[MAXEDGE];
	int inq[MAXNODE];
	Type d[MAXNODE];
	int p[MAXNODE];
	Type a[MAXNODE];

	void init(int n) {
		this->n = n;
		memset(first, -1, sizeof(first));
		m = 0;
	}

	void add_Edge(int u, int v, Type cap, Type cost) {
		edges[m] = Edge(u, v, cap, 0, cost);
		next[m] = first[u];
		first[u] = m++;
		edges[m] = Edge(v, u, 0, 0, -cost);
		next[m] = first[v];
		first[v] = m++;
	}

	bool bellmanford(int s, int t, Type &flow, Type &cost) {

		for (int i = 0; i < n; i++) d[i] = INF;
		memset(inq, false, sizeof(inq));
		d[s] = 0; inq[s] = true; p[s] = s; a[s] = INF;
		queue<int> Q;
		Q.push(s);
		while (!Q.empty()) {
			int u = Q.front(); Q.pop();
			inq[u] = false;
			for (int i = first[u]; i != -1; i = next[i]) {
				Edge& e = edges[i];
				if (e.cap > e.flow && d[e.v] > d[u] + e.cost) {
					d[e.v] = d[u] + e.cost;
					p[e.v] = i;
					a[e.v] = min(a[u], e.cap - e.flow);
					if (!inq[e.v]) {Q.push(e.v); inq[e.v] = true;}
				}
			}
		}
		if (d[t] == INF) return false;
		flow += a[t];
		cost += d[t] * a[t];
		int u = t;
		while (u != s) {
			edges[p[u]].flow += a[t];
			edges[p[u]^1].flow -= a[t];
			u = edges[p[u]].u;
		}
		return true;
	}

	Type Mincost(int s, int t, Type sum) {
		Type flow = 0, cost = 0;
		while (bellmanford(s, t, flow, cost));
		if (sum != flow) cost = -1;
		return cost;
	}
} gao;

const int N = 55;

int n, m, k, sum;
int a[N][N], b[N][N];

int solve() {
	int c, ans = 0, flag = 0;
	for (int i = 0; i < k; i++) {
		gao.init(n + m + 2);
		for (int j = 1; j <= m; j++)
			gao.add_Edge(0, j, b[j][i], 0);
		int sum = 0;
		for (int j = 1; j <= n; j++) {
			sum += a[j][i];
			gao.add_Edge(m + j, n + m + 1, a[j][i], 0);
		}
		for (int x = 1; x <= n; x++) {
			for (int y = 1; y <= m; y++) {
				scanf("%d", &c);
				gao.add_Edge(y, x + m, INF, c);
			}
		}
		if (flag) continue;
		int tmp = gao.Mincost(0, n + m + 1, sum);
		if (tmp == -1) flag = 1;
		else ans += tmp;
	}
	if (flag) ans = -1;
	return ans;
}

int main() {
	while (~scanf("%d%d%d", &n, &m, &k) && n) {
		for (int i = 1; i <= n; i++)
			for (int j = 0; j < k; j++)
				scanf("%d", &a[i][j]);
		for (int i = 1; i <= m; i++)
			for (int j = 0; j < k; j++)
				scanf("%d", &b[i][j]);
		printf("%d\n", solve());
	}
	return 0;
}

POJ 2516 Minimum Cost(费用流）