分配问题 费用流

题目描述

有 nnn 件工作要分配给 nnn 个人做。第 iii 个人做第 jjj 件工作产生的效益为 cijc_{ij}cij​ 。试设计一个将 nnn 件工作分配给 nnn 个人做的分配方案，使产生的总效益最大。

输入输出格式

输入格式：

文件的第 111 行有 111 个正整数 nnn，表示有 nnn 件工作要分配给 nnn 个人做。

接下来的 nnn 行中，每行有 nnn 个整数 cijc_{ij}cij​​​，表示第 iii 个人做第 jjj 件工作产生的效益为 cijc_{ij}cij​。

输出格式：

两行分别输出最小总效益和最大总效益。

输入输出样例

输入样例#1： 复制

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

输出样例#1： 复制

5
14

说明

1≤n≤1001 leq n leq 1001≤n≤100

一个人只能修一个工件

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<time.h>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 20005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;

inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}


ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }



/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

bool vis[maxn];
int n, m, s, t;
int x, y, f, z;
int dis[maxn], pre[maxn], last[maxn], flow[maxn];
int maxflow, mincost;

struct node {
	int to, nxt, flow, dis;
}edge[maxn << 2];

int head[maxn], cnt;


void addedge(int from, int to, int flow, int dis) {
	edge[++cnt].to = to; edge[cnt].flow = flow; edge[cnt].dis = dis;
	edge[cnt].nxt = head[from]; head[from] = cnt;
}

bool spfa1(int s, int t) {
	memset(dis, 0x7f, sizeof(dis)); 
	memset(flow, 0x7f, sizeof(flow));
	ms(vis);queue<int>q;
	q.push(s); vis[s] = 1; dis[s] = 0; pre[t] = -1;
	while (!q.empty()) {
		int now = q.front(); q.pop(); vis[now] = 0;
		for (int i = head[now]; i != -1; i = edge[i].nxt) {
			if (edge[i].flow > 0 && dis[edge[i].to] > dis[now] + edge[i].dis) {
				dis[edge[i].to] = edge[i].dis + dis[now];
				pre[edge[i].to] = now; last[edge[i].to] = i;
				flow[edge[i].to] = min(flow[now], edge[i].flow);
				if (!vis[edge[i].to]) {
					vis[edge[i].to] = 1; q.push(edge[i].to);
				}
			}
		}
	}
	return pre[t] != -1;
}
bool spfa2(int s, int t) {
	memset(dis, 0x7f, sizeof(dis));
	memset(flow, 0x7f, sizeof(flow));
	ms(vis); queue<int>q;
	q.push(s); vis[s] = 1; dis[s] = 0; pre[t] = -1;
	while (!q.empty()) {
		int now = q.front(); q.pop(); vis[now] = 0;
		for (int i = head[now]; i != -1; i = edge[i].nxt) {
			if (edge[i].flow > 0 && dis[edge[i].to] > dis[now] + edge[i].dis) {
				dis[edge[i].to] = edge[i].dis + dis[now];
				pre[edge[i].to] = now; last[edge[i].to] = i;
				flow[edge[i].to] = min(flow[now], edge[i].flow);
				if (!vis[edge[i].to]) {
					vis[edge[i].to] = 1; q.push(edge[i].to);
				}
			}
		}
	}
	return pre[t] != -1;
}

void mincost_maxflow() {
	while (spfa1(s, t)) {
		int now = t;
		maxflow += flow[t]; mincost += flow[t] * dis[t];
		while (now != s) {
			edge[last[now]].flow -= flow[t];
			edge[last[now] ^ 1].flow += flow[t];
			now = pre[now];
		}
	}
}
void maxcost_maxflow() {
	while (spfa2(s, t)) {
		int now = t;
		maxflow += flow[t]; mincost += flow[t] * dis[t];
		while (now != s) {
			edge[last[now]].flow -= flow[t];
			edge[last[now] ^ 1].flow += flow[t];
			now = pre[now];
		}
	}
}

int C[104][104];

int main()
{
	//	ios::sync_with_stdio(0);
	n = rd(); mclr(head, -1); cnt = 1;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++)C[i][j] = rd();
	}
	s = 0; t = 2 * n + 2;
	for (int i = 1; i <= n; i++) {
		addedge(s, i, 1, 0); addedge(i, s, 0, 0);
	}
	for (int i = 1; i <= n; i++) {
		addedge(i + n, t, 1, 0); addedge(t, i + n, 0, 0);
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			addedge(i, j + n, 1, C[i][j]); addedge(j + n, i, 0, -C[i][j]);
		}
	}
	int ans1 = 0, ans2 = 0;
	mincost_maxflow(); ans1 = mincost;
	mclr(head, -1); cnt = 1;
	ms(edge); ms(pre); ms(last); mincost = maxflow = 0;
	for (int i = 1; i <= n; i++) {
		addedge(s, i, 1, 0); addedge(i, s, 0, 0);
	}
	for (int i = 1; i <= n; i++) {
		addedge(i + n, t, 1, 0); addedge(t, i + n, 0, 0);
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			addedge(i, j + n, 1, -C[i][j]); addedge(j + n, i, 0, C[i][j]);
		}
	}
	maxcost_maxflow();
	ans2 = mincost;
	cout << ans1 << endl; cout << -ans2 << endl;
	return 0;
}