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

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
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.

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

题意：n个城市，m条边， 每个点刚开始有ai个人，问最后能不能有bi个人，每个城市的人，要么不走，要么只能走到相邻的城市

思路：最大流 s-a[i]-b[i]-t建图

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
const int maxn=5000;
const int  INF=1000000000;
int N,M;
int A[maxn],B[maxn];
int tot,head[maxn];
int st,en,nn,cur[maxn];
int ans[maxn][maxn];
int dis[maxn],gap[maxn],vis[maxn],pre[maxn];
struct node
{
    int v,next,f,flow;
}edge[maxn*4];
void init()
{
    tot=0;
    memset(head,-1,sizeof(head));
}
void add_edge(int x,int y,int f)
{
    edge[tot].v=y;
    edge[tot].f=f;
    edge[tot].flow=0;
    edge[tot].next=head[x];
    head[x]=tot++;
    edge[tot].v=x;
    edge[tot].f=0;
    edge[tot].flow=0;
    edge[tot].next=head[y];
    head[y]=tot++;
}
int SAP(int st,int en)
{
    for(int i=0;i<=nn;i++)
    {
        cur[i]=head[i];
        dis[i]=gap[i]=0;
    }
    int u=0;
    int flow=0,aug=INF;
    gap[st]=nn;
    u=pre[st]=st;
    bool flag;
    while(dis[st]<nn)
    {
        flag=0;
        for(int &j=cur[u];j!=-1;j=edge[j].next)
        {
            int v=edge[j].v;
            if(edge[j].f>0&&dis[u]==dis[v]+1)
            {
                flag=1;
                if(edge[j].f<aug)aug=edge[j].f;
                pre[v]=u;
                u=v;
                if(u==en)
                {
                    flow+=aug;
                    while(u!=st)
                    {
                        u=pre[u];
                        edge[cur[u]].f-=aug;
                        edge[cur[u]].flow-=aug;
                        edge[cur[u]^1].f+=aug;
                        edge[cur[u]^1].flow+=aug;
                    }
                    aug=INF;
                }
                break;
            }
        }

        if(flag)continue;
        int mindis=nn;
        for(int j=head[u];j!=-1;j=edge[j].next)
        {
            int v=edge[j].v;
            if(dis[v]<mindis&&edge[j].f>0)
            {
                mindis=dis[v];
                cur[u]=j;
            }
        }
        if((--gap[dis[u]])==0)break;
        gap[dis[u]=mindis+1]++;
        u=pre[u];
    }
    return flow;
}
int main()
{
    scanf("%d%d",&N,&M);
    st=0,en=2*N+1,nn=en+1;
    init();
    int sum=0,sum1=0;
    for(int i=1;i<=N;i++)
    {
        scanf("%d",&A[i]);
        sum+=A[i];
        add_edge(st,i,A[i]);
    }
    for(int i=1;i<=N;i++)
    {
        scanf("%d",&B[i]);
        sum1+=B[i];
        add_edge(i,i+N,INF);
        add_edge(i+N,en,B[i]);
    }
    while(M--)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        add_edge(u,v+N,INF);
        add_edge(v,u+N,INF);
    }

    if(sum!=sum1){printf("NO\n");return 0;}
    int flow=SAP(st,en);
    if(flow!=sum){printf("NO\n");}
    else
    {
        for(int i=1;i<=N;i++)
            for(int j=head[i];j!=-1;j=edge[j].next)
            {
                int v=edge[j].v;
                ans[i][v-N]=-edge[j].flow;
            }
        printf("YES\n");
        for(int i=1;i<=N;i++)
        {
            for(int j=1;j<=N;j++)
                printf("%d ",ans[i][j]);
            printf("\n");
        }
    }
    return 0;
}