E. Soldier and Traveling (CF 304 div2) 最大流

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 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.

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：给一个无向图，每个点有一个权，这是初态，给出目标态，问能否达到目标态，只能向相邻的点传递权。

思路：网络流，s->a[i]->b[i]->t建图。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define MAXN 205
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
const int MAXM = 2010;
typedef long long ll;
using namespace std;

struct Edge
{
    int to,next,cap,flow;
}edge[MAXM];

int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];

void init()
{
    tol=0;
    memset(head,-1,sizeof(head));
}

//加边，单向图三个参数，双向图四个参数
void addedge(int u,int v,int w,int rw=0)
{
    edge[tol].to=v; edge[tol].cap=w; edge[tol].next=head[u];
    edge[tol].flow=0; head[u]=tol++;
    edge[tol].to=u; edge[tol].cap=rw; edge[tol].next=head[v];
    edge[tol].flow=0; head[v]=tol++;
}

//输入参数：起点，终点，点的总数
//点的编号没有影响，只要输入点的总数
int sap(int start,int end,int N)
{
    memset(gap,0,sizeof(gap));
    memset(dep,0,sizeof(dep));
    memcpy(cur,head,sizeof(head));
    int u=start;
    pre[u]=-1;
    gap[0]=N;
    int ans=0;
    while (dep[start]<N)
    {
        if (u==end)
        {
            int Min=INF;
            for (int i=pre[u];i!=-1;i=pre[edge[i^1].to])
                if (Min>edge[i].cap-edge[i].flow)
                    Min=edge[i].cap-edge[i].flow;
            for (int i=pre[u];i!=-1;i=pre[edge[i^1].to])
            {
                edge[i].flow+=Min;
                edge[i^1].flow-=Min;
            }
            u=start;
            ans+=Min;
            continue;
        }
        bool flag=false;
        int v;
        for (int i=cur[u];i!=-1;i=edge[i].next)
        {
            v=edge[i].to;
            if (edge[i].cap-edge[i].flow && dep[v]+1==dep[u])
            {
                flag=true;
                cur[u]=pre[v]=i;
                break;
            }
        }
        if (flag)
        {
            u=v;
            continue;
        }
        int Min=N;
        for (int i=head[u];i!=-1;i=edge[i].next)
            if (edge[i].cap-edge[i].flow && dep[edge[i].to]<Min)
            {
                Min=dep[edge[i].to];
                cur[u]=i;
            }
        gap[dep[u]]--;
        if (!gap[dep[u]]) return ans;
        dep[u]=Min+1;
        gap[dep[u]]++;
        if (u!=start) u=edge[pre[u]^1].to;
    }
    return ans;
}

int mp[MAXN][MAXN];
int n,m;
int a[MAXN],b[MAXN];

int main()
{
#ifndef ONLINE_JUDGE
    freopen("C:/Users/asus1/Desktop/IN.txt","r",stdin);
#endif
    int i,j,u,v;
    while (~sff(n,m))
    {
        init();
        int s1=0,s2=0;
        for (i=1;i<=n;i++)
        {
            sf(a[i]);
            s1+=a[i];
            addedge(0,i,a[i]);
        }
        for (i=1;i<=n;i++)
        {
            sf(b[i]);
            s2+=b[i];
            addedge(i,i+n,INF);
            addedge(i+n,2*n+1,b[i]);
        }
        for (i=0;i<m;i++)
        {
            sff(u,v);
            addedge(u,v+n,INF);
            addedge(v,u+n,INF);
        }
        if (s1!=s2){
            pf("NO\n");
            continue;
        }
        if (sap(0,2*n+1,2*n+2)!=s1){
            pf("NO\n");
            continue;
        }
        pf("YES\n");
        memset(mp,0,sizeof(mp));
        for (i=1;i<=n;i++)
        {
            for (j=head[i];j+1;j=edge[j].next)
            {
                if (edge[j].to==0) continue;
                mp[i][edge[j].to-n]=edge[j].flow;
            }
        }
        for (i=1;i<=n;i++)
        {
            for (j=1;j<=n;j++)
                pf("%d ",mp[i][j]);
            pf("\n");
        }
    }
    return 0;
}