Soldier and Traveling （网络流最大流）

我不过是个神秘人

于 2019-07-13 15:48:41 发布

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本文链接：https://blog.youkuaiyun.com/weixin_43326028/article/details/95757404

博客围绕国家中城市士兵移动问题展开。给定城市数量、道路连接情况、各城市初始和目标士兵数量，需判断能否通过士兵移动达到目标状态。输入包含城市与道路信息、初始和目标士兵数，输出判断结果及移动方案。

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In the country there are n cities and m bidirectional roads between them. Each city has an army. Army of the i-th city consists of ai soldiers. Now soldiers roam. After roaming each soldier has to either stay in his city or to go to the one of neighboring cities by at moving along at most one road.

Check if is it possible that after roaming there will be exactly bi soldiers in the i-th city.

Input
First line of input consists of two integers n and m (1 ≤ n ≤ 100, 0 ≤ m ≤ 200).

Next line contains n integers a1, a2, …, an (0 ≤ ai ≤ 100).

Next line contains n integers b1, b2, …, bn (0 ≤ bi ≤ 100).

Then m lines follow, each of them consists of two integers p and q (1 ≤ p, q ≤ n, p ≠ q) denoting that there is an undirected road between cities p and q.

It is guaranteed that there is at most one road between each pair of cities.

Output
If the conditions can not be met output single word “NO”.

Otherwise output word “YES” and then n lines, each of them consisting of n integers. Number in the i-th line in the j-th column should denote how many soldiers should road from city i to city j (if i ≠ j) or how many soldiers should stay in city i (if i = j).

If there are several possible answers you may output any of them.

Examples
Input
4 4
1 2 6 3
3 5 3 1
1 2
2 3
3 4
4 2
Output
YES
1 0 0 0
2 0 0 0
0 5 1 0
0 0 2 1
Input
2 0
1 2
2 1
Output
NO

例一的网络流的图如下，自身到自身为i->i+n
在这里插入图片描述

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<vector>
#include<cmath>
#define LL long long 
#define mem(a, b) memset(a, b, sizeof(a))
#define N 210 
#define MOD
using namespace std;
const int inf=1<<29;
int n, m, T;
int head[N], ver[2001], edge[2001], Next[2001], v[N], incf[N], pre[N];
int s, t, tot, maxflow, a[N], b[N], sum1, sum2, ans[N][N];

void add(int x, int y, int z) {
	ver[++tot]=y, edge[tot]=z, Next[tot]=head[x], head[x]=tot;
	ver[++tot]=x, edge[tot]=0, Next[tot]=head[y], head[y]=tot;
}

bool bfs() {
	mem(v, 0);
	queue<int> q;
	q.push(s); v[s]=1;
	incf[s]=inf;
	while(q.size()) {
		int x=q.front(); q.pop();
		for(int i=head[x]; i; i=Next[i])
			if(edge[i]) {
				int y=ver[i];
				if(v[y]) continue;
				incf[y]=min(incf[x], edge[i]);
				pre[y]=i;
				q.push(y), v[y]=1;
				if(y==t) return 1;
			}
	}
	return 0;
}

void update() {
	int x=t;
	while(x!=s) {
		int i=pre[x];
		edge[i]-=incf[t];
		edge[i^1]+=incf[t];
		x=ver[i^1]; 
	}
	maxflow+=incf[t];
}

int main() {
	while(scanf("%d%d",&n, &m)!=EOF) {	
		sum1=0, sum2=0;
		mem(ans, 0); 
		maxflow=0;
		mem(head, 0);
		s=0, t=2*n+1;
		tot=1;
		for(int i=1; i<=n; i++) {
			scanf("%d", &a[i]);
			sum1+=a[i];
			add(s, i, a[i]);
			add(i, i+n, inf); 
		}
		for(int i=1; i<=n; i++){
			scanf("%d",&b[i]);
			sum2+=b[i];
			add(i+n, t, b[i]);
		} 
		for(int i=1; i<=m; i++) {
			int x, y;
			scanf("%d%d", &x, &y);
			add(x, y+n, inf);
			add(y, x+n, inf); 
		}
		while(bfs()) update();
		if(sum1==sum2 && sum1==maxflow) {
			cout<<"YES\n";
			for(int i=1; i<=n; i++) {
				for(int j=head[i]; j; j=Next[j]) {
					int x=ver[j]-n;
					if(x>=1 && x<=n && edge[j^1]>0) ans[i][x]=edge[j^1];
				}
			}
			for(int i=1; i<=n; i++) {
				for(int j=1; j<=n; j++) {
					cout<<ans[i][j]<<" ";
				}
				cout<<endl;
			}
		}
		else
			cout<<"NO\n";
	}
}