「网络流 24 题」分配问题

题目描述

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 cij c_{ij}c​ij​​
。

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。

输入格式

文件的第 $1$ 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 c_{ij}c​ij​​，表示第 $i$ 个人做第 $j$ 件工作产生的效益为 cij c_{ij}c​ij​​。

输出格式

两行分别输出最小总效益和最大总效益。

样例

样例输入

52 2 2 1 22 3 1 2 42 0 1 1 12 3 4 3 33 2 1 2 1

样例输出

514

数据范围与提示

$1\le n\le 100$

#include<stdio.h>#include<algorithm>#include<string.h>#include<queue>using namespace std;const int maxm = 10005;const int maxn = 100005;const int INF = 1e9 + 7;struct node{int u, v, flow, cost, next;}edge[maxn];int dis[maxm], head[maxm], cur[maxm], pre[maxn], f[1005][1005], map[1005][1005], a[maxm], b[maxm];int s, t, n, m, cnt;void init(){cnt = 0, s = 0, t = 10000;memset(head, -1, sizeof(head));}void add(int u, int v, int w, int cost){edge[cnt].u = u, edge[cnt].v = v;edge[cnt].flow = w, edge[cnt].cost = cost;edge[cnt].next = head[u], head[u] = cnt++;edge[cnt].u = v, edge[cnt].v = u;edge[cnt].flow = 0, edge[cnt].cost = -cost;edge[cnt].next = head[v], head[v] = cnt++;}int bfs(){queue<int>q;for (int i = 0;i <= 10004;i++) dis[i] = INF;memset(pre, -1, sizeof(pre));dis[s] = 0;q.push(s);int rev = 0;while (!q.empty()){int u = q.front();q.pop();rev++;for (int i = head[u];i != -1;i = edge[i].next){int v = edge[i].v;if (dis[v] > dis[u] + edge[i].cost&&edge[i].flow){dis[v] = dis[u] + edge[i].cost;pre[v] = i;q.push(v);}}}if (dis[t] == INF) return 0;return 1;}int MCMF(){int minflow, ans = 0;while (bfs()){minflow = INF;for (int i = pre[t];i != -1;i = pre[edge[i].u])minflow = min(minflow, edge[i].flow);for (int i = pre[t];i != -1;i = pre[edge[i].u]){edge[i].flow -= minflow;edge[i ^ 1].flow += minflow;}ans += minflow*dis[t];}return ans;}int main(){int i, j, k, sum, id = 0;scanf("%d", &n);init();for (i = 1;i <= n;i++)add(i, t, 1, 0);for (i = 1;i <= n;i++){add(s, i + n, 1, 0);for (j = 1;j <= n;j++){scanf("%d", &f[i][j]);add(i + n, j, 1, f[i][j]);}}printf("%d\n", MCMF());init();for (i = 1;i <= n;i++)add(i, t, 1, 0);for (i = 1;i <= n;i++){add(s, i + n, 1, 0);for (j = 1;j <= n;j++)add(i + n, j, 1, -f[i][j]);}printf("%d\n", -MCMF());return 0;}