Codeforces Round #304 (Div. 2)-E. Soldier and Traveling

原题链接

E. Soldier and Traveling

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 aisoldiers. 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

这道题就是网络流，0为源点，和1到n城市各连一条边，边的容量为城市中士兵的数量，若两个城市i, j之间有边，则i, j+n和j, i+n各连一条边，在从n+1..2n与2n+1连边，

边的容量为各个城市最终的士兵数量

#include <iostream>
#include <cstdio> 
#include <algorithm>
#include <cstring>
#include <vector>
#include <map>
#include <set>
#include <queue>
#define maxn 100005
#define INF 1000000009
#define MOD 1000000007
typedef long long ll;
using namespace std;

struct Edge{
	Edge(){
	}
	Edge(int a, int b, int c, int d){
		from = a;
		to = b;
		flow = c;
		cap = d;
	}
	int from, to, flow, cap;
};
vector<Edge> edge;
vector<int> v[305];
int n, m, a[105], ans[105][105], vis[305], pre[305], sum, sum2;
void Add(int k1, int k2, int p){
	edge.push_back(Edge(k1, k2, 0, p));
	edge.push_back(Edge(k2, k1, 0, 0));
	int s = edge.size();
	v[k1].push_back(s-2);
	v[k2].push_back(s-1);
}
bool MaxFlow(){
	int ss = 0;
	while(1){
		queue<int> q;
		q.push(0);
		memset(vis, 0, sizeof(vis));
		vis[0] = 1e9;
		pre[0] = -1;
		while(!q.empty()){
			int h = q.front();
			q.pop();
			for(int i = 0; i < v[h].size(); i++){
				Edge e = edge[v[h][i]];
				if(vis[e.to] == 0 && e.cap > e.flow){
					vis[e.to] = min(vis[e.from], e.cap - e.flow);
					q.push(e.to);
					//printf("%d\n", e.to);
					pre[e.to] = v[h][i];
				}
			}
			if(vis[2*n+1])
			 break;
		}
		if(vis[2*n+1] == 0)
		 break;
		ss += vis[2*n+1];
		for(int j = pre[2*n+1]; j != -1; j = pre[edge[j].from]){
			edge[j].flow += vis[2*n+1];
			edge[j^1].flow -= vis[2*n+1];
		}
    }
	if(ss == sum)
	 return true;
	return false;
}
int main(){
//	freopen("in.txt", "r", stdin);
	int k1, k2;
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i++){
		scanf("%d", a+i);
		sum += a[i];
	}
	for(int i = 1; i <= n; i++){
		scanf("%d", &k1);
		Add(i+n, 2*n+1, k1);
		sum2 += k1;
	}
	if(sum != sum2){
		puts("NO");
		return 0;
	}
	for(int i = 0; i < m; i++){
		scanf("%d%d", &k1, &k2);
		Add(k1, k2+n, a[k1]);
		Add(k2, k1+n, a[k2]);
	}
	for(int i = 1; i <= n; i++){
	 Add(0, i, a[i]);
	 Add(i, i+n, a[i]);
    }
	if(MaxFlow()){
		puts("YES");
		for(int i = 1; i <= n; i++){
			for(int j = 0; j < v[i].size(); j++){
				int h = v[i][j];
				if(edge[h].to){
					ans[i][edge[h].to-n] += edge[h].flow;
			    }
			}
		}
		for(int i = 1; i <= n; i++){
		 for(int j = 1; j <= n; j++){
		 	if(j > 1)
		 	 putchar(' ');
		 	printf("%d", ans[i][j]);
		 }
		 puts("");
	   }
	}
	else{
		puts("NO");
	}
	return 0;
}