codeforces 546 E. Soldier and Traveling

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 atmoving 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.

Sample test(s)

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

网络流，然后利用残余网络统计花掉的流量，建图参考代码。

/*======================================================
# Author: whai
# Last modified: 2015-10-23 16:08
# Filename: e.cpp
======================================================*/
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#include <set>
#include <map>
#include <queue>
using namespace std;

#define LL __int64
#define PB push_back
#define P pair<int, int>
#define X first
#define Y second

const int N = 105;

int a
, b
;

struct Edge {
int to, cap, rev;
};

const int MAX_V = 405;
const int INF = 0x3f3f3f3f;

vector<Edge> G[MAX_V];

int level[MAX_V];
int iter[MAX_V];

void add_edge(int from, int to, int cap) {
G[from].PB((Edge) {to, cap, G[to].size()});
G[to].PB((Edge) {from, 0, G[from].size() - 1});
}

void bfs(int s) {
memset(level, -1, sizeof(level));
queue<int> que;
level[s] = 0;
que.push(s);
while (!que.empty()) {
int v = que.front(); que.pop();
for(int i = 0; i < G[v].size(); ++i) {
Edge &e = G[v][i];
if(e.cap > 0 && level[e.to] < 0) {
level[e.to] = level[v] + 1;
que.push(e.to);
}
}
}
}

int dfs(int v, int t, int f) {
if(v == t) return f;
for(int &i = iter[v]; i < G[v].size(); ++i) {
Edge &e = G[v][i];
if (e.cap > 0 && level[v] < level[e.to]) {
int d = dfs(e.to, t, min(f, e.cap));
if (d > 0) {
e.cap -= d;
G[e.to][e.rev].cap += d;
return d;
}
}
}
return 0;
}

int max_flow(int s, int t) {
int flow = 0;
while (1) {
bfs(s);
if (level[t] < 0) return flow;
memset(iter, 0, sizeof(iter));
int f;
while ((f = dfs(s, t, INF)) > 0) {
flow += f;
}
}
return flow;
}

int ans

;

bool used

;

int main() {
int n, m;
scanf("%d%d", &n, &m);
int sum0 = 0;
for(int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
sum0 += a[i];
}

int sum1 = 0;
for(int i = 1; i <= n; ++i) {
scanf("%d", &b[i]);
sum1 += b[i];
}

for(int i = 1; i <= n; ++i) {
add_edge(0, i, a[i]);
add_edge(i, i + n, a[i]);
used[i][i] = 1;
ans[i - 1][i - 1] = a[i];
add_edge(i + n, 2 * n + 1, b[i]);
}

for(int i = 0; i < m; ++i) {
int u, v;
scanf("%d%d", &u, &v);
if(used[u][v]) continue;
used[u][v] = used[v][u] = 1;
ans[u - 1][v - 1] = a[u];
ans[v - 1][u - 1] = a[v];
add_edge(u, v + n, a[u]);
add_edge(v, u + n, a[v]);
}
//for(int i = 0; i < n; ++i) {
//	for(int j = 0; j < n; ++j) {
//		cout<<ans[i][j]<<' ';
//	}
//	cout<<endl;
//}

int flow = max_flow(0, 2 * n + 1);
//cout<<flow<<endl;
if(sum0 == sum1 && flow == sum0) {
cout<<"YES"<<endl;
for(int i = 1; i <= n; ++i) {
for(int j = 0; j < G[i].size(); ++j) {
Edge e = G[i][j];
int u = i - 1;
int v = e.to;
if(v > n) v = v - n - 1;
else v = v - 1;
ans[u][v] -= e.cap;
}
}
for(int i = 0; i < n; ++i) {
for(int j = 0; j < n; ++j) {
cout<<ans[i][j]<<' ';
}
cout<<endl;
}

} else {
cout<<"NO"<<endl;
}

return 0;
}