CODEVS 1915 分配问题

题目描述 Description

有n件工作要分配给n个人做。第i 个人做第j 件工作产生的效益为ij c 。试设计一个将

n件工作分配给n个人做的分配方案,使产生的总效益最大。

«编程任务:

对于给定的n件工作和n个人,计算最优分配方案和最差分配方案。

输入描述 Input Description

第1 行有1 个正整数n,表示有n件工作要分配给n 个人做。接下来的n 行中,每行有n 个整数 cij ,1≤i≤n,1≤j≤n,表示第i 个人做第j件工作产生的效益为cij

输出描述 Output Description

将计算出的最小总效益和最大总效益输出

样例输入 Sample Input

5

2 2 2 1 2

2 3 1 2 4

2 0 1 1 1

2 3 4 3 3

3 2 1 2 1

样例输出 Sample Output

5

14

Solution

这道题可以KM做,为了联系下网络流,用MCMF做的,跑一边min cost max flow, 还有max cost max flow就可以了。

Code

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#include <cstdio>
#include <queue>
using namespace std;
int getint() {
	int r = 0, k = 1; char c = getchar();
	for (; c < '0' || c > '9'; c = getchar()) if (c == '-') k = -1;
	for (; '0' <= c && c <= '9'; c = getchar()) r = r * 10 - '0' + c;
	return r * k;
}
const int maxn = 105;
const int maxe = maxn*maxn+2*maxn+10;
const int maxnode = maxn*2+10;
int n, S, T;
struct edge_type {
	int to, next, r, c;
} edge[maxe<<1];
int cnte = 1, h[maxnode];
void ins(int u, int v, int f, int c) {
	edge[++cnte].to = v;
	edge[cnte].r = f;
	edge[cnte].c = c;
	edge[cnte].next = h[u];
	h[u] = cnte;
	edge[++cnte].to = u;
	edge[cnte].r = 0;
	edge[cnte].c = -c;
	edge[cnte].next = h[v];
	h[v] = cnte;
}

queue<int> Q;
bool vis[maxnode];
int dis[maxnode], w[maxn][maxn], pre[maxnode];
int a[maxnode];
const int INF = 1<<30;

bool SPFA(int S, int T, int &flow, int &cost) {
	while (!Q.empty()) Q.pop();
	for (int i = S; i <= T; ++i) {
		vis[i] = false;
		dis[i] = INF;
	}
	vis[S] = true; dis[S] = 0; a[S] = INF; pre[S] = 0;
	Q.push(S);
	int now;
	while (!Q.empty()) {
		now = Q.front(); Q.pop(); vis[now] = false;
		for (int i = h[now]; i; i = edge[i].next) {
			if (edge[i].r && dis[edge[i].to] > dis[now] + edge[i].c) {
				dis[edge[i].to] = dis[now] + edge[i].c;
				a[edge[i].to] = min(a[now], edge[i].r);
				pre[edge[i].to] = i;
				if (!vis[edge[i].to]) {
					vis[edge[i].to] = true;
					Q.push(edge[i].to);
				}
			}
		}
	}
	if (dis[T] == INF) return false;
	flow += a[T];
	cost += a[T] * dis[T];
	for (now = T; now != S; now = edge[pre[now]^1].to) {
		edge[pre[now]].r -= a[T];
		edge[pre[now]^1].r += a[T];
	}
	return true;
}
int MCMF(int S, int T) {
	int flow = 0, cost = 0;
	while (SPFA(S, T, flow, cost));
	return cost;
}
int main() {
	n = getint();
	S = 1; T = 2*n+2;
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= n; ++j) {
			w[i][j] = getint();
			ins(i+1, j+n+1, 1, w[i][j]);
		}
		ins(S, i+1, 1, 0);
		ins(i+n+1, T, 1, 0);
	}
	int ans = MCMF(S, T);
	printf("%d\n", ans);
	cnte = 1;
	for (int i = S; i <= T; ++i) h[i] = 0;
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= n; ++j) {
			ins(i+1, j+n+1, 1, -w[i][j]);
		}
		ins(S, i+1, 1, 0);
		ins(i+n+1, T, 1, 0);
	}
	ans = MCMF(S, T);
	printf("%d", -ans);
}