【codeforces 546E】Soldier and Traveling

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard 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 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

【题目链接】:http://codeforces.com/problemset/problem/546/E

【题解】



题意:

每个点上的士兵只能走到相邻的点(只能走一次);

然后问你目标状态能不能达到;

做法:

按照下面这张图的规则建边;

把n个点每个点分成入点和出点两个点;

虚拟一个起点和终点(作为源点和汇点);

入点都和起点连在一起;边权为ai;

出点都和终点连在一起；边权为bi;

然后入点和出点之间接一条边;边权为INF;

如果x和y之间有一条边;

则在x的入点和y的出点之间,以及y的入点以及x的出点之间建INF的边;

然后从起点开始跑最大流;



这样造成的效果是,每个点的士兵都只会到和其距离为1的点（然后进入汇点)

显然∑ai==∑bi要满足

这样f[起点][1..i]和f[i+n][汇点]必然都是满流的;

以这个作为判断是否有解的依据;

然后输出每个点到其他点的流量就好;



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int MAXN = 100+10;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

int n,m,s,t;
int g[MAXN*2][MAXN*2],f[MAXN*2][MAXN*2];
int flag[MAXN*2];

int dfs(int x,int m)
{
    if (x==t || !m) return m;
    if (flag[x]++) return 0;
    rep1(y,1,2*n+2)
    {
        int judge = dfs(y,min(m,g[x][y]-f[x][y]));
        if (judge)
        {
            f[x][y]+=judge;
            f[y][x]-=judge;
            return judge;
        }
    }
    return 0;
}

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    rei(n);rei(m);
    s = 2*n+1,t = 2*n+2;
    rep1(i,1,n)
        {
            int x;
            rei(x);
            g[s][i] = x;
        }
    rep1(i,1,n)
        {
            int x;
            rei(x);
            g[i+n][t] = x;
        }
    rep1(i,1,n)
        g[i][i+n] = 100000;
    rep1(i,1,m)
    {
        int x,y;
        rei(x);rei(y);
        g[x][y+n] = 100000;
        g[y][x+n] = 100000;
    }
    while (dfs(s,100000))
        memset(flag,0,sizeof flag);
    rep1(i,1,n)
        if (f[s][i]!=g[s][i] || f[i+n][t]!=g[i+n][t])
        {
            puts("NO");
            return 0;
        }
    puts("YES");
    rep1(i,1,n)
        rep1(j,1,n)
        {
            printf("%d",f[i][j+n]);
            if (j==n)
                puts("");
            else
                putchar(' ');
        }
    return 0;
}