【codeforces 546E】Soldier and Traveling
2016-12-28       个评论    来源：AWCXV
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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

【题解】

vcbky/u147XEwffBv77NusM7PGJyIC8+DQo8YnIgLz4NCqG+zerV+7T6wuuhvzwvcD4NCjxwcmUgY2xhc3M9"brush:java;"> #include 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 pii; typedef pair 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; }