codeforces 546 E. Soldier and Traveling(網絡流＋路徑輸出)

題目鏈接：https://codeforces.com/problemset/problem/546/E

思路：很簡單的建圖，留個輸出路徑的板子

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%I64d",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(998244353)
#define pb push_back
#define eps 1e-6
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
const int N=1008,M=2008;
int head[N],tot,n,m,x,y,a[N],b[N];
struct node{
int next,c,to;}e[M];
struct max_flow{
    int S,T,n;
    int lev[N],q[N],cur[N],f;
    void init(int _s,int _t)
    {
        tot=0;S=_s,T=_t,n=T+1;
        FOR(i,0,n) head[i]=-1;
    }
    void add(int a,int b,int c)
    {
        e[tot].next=head[a];
        e[tot].to=b;
        e[tot].c=c;
        head[a]=tot++;
    }
    void Add(int a,int b,int c)
    {
        add(a,b,c);
        add(b,a,0);
    }
    int bfs()
    {
        FOR(i,0,n) lev[i]=0;
        lev[S]=1,f=0,q[f++]=S;
        FOR(i,0,f-1)
        {
            int u=q[i];
            for(int i=head[u];i!=-1;i=e[i].next)
             if(lev[e[i].to]==0&&e[i].c>0)
             {
                 int to=e[i].to;
                 lev[to]=lev[u]+1;
                 q[f++]=to;
                 if(to==T) return 1;
             }
        }
        return 0;
    }
    int dfs(int u,int f)
    {
        if(u==T) return f;
        int tag=0,c;
        for(int &i=cur[u];i!=-1;i=e[i].next)
        {
            int to=e[i].to;
            if(e[i].c>0&&lev[to]==lev[u]+1)
            {
                c=dfs(to,min(f-tag,e[i].c));
                e[i].c-=c;
                e[i^1].c+=c;
                tag+=c;
                if(tag==f) return tag;
            }
        }
        return tag;
    }
    int slove()
    {
        int ans=0;
        while(bfs())
        {
            FOR(i,0,n) cur[i]=head[i];
            ans+=dfs(S,inf);
        }
        return ans;
    }
    int mat[N][N];
    void print(int n)
    {
        REW(mat,0);
        FOR(i,1,n)
         for(int j=head[i+n];j!=-1;j=e[j].next)
          if(e[j].c!=inf&&e[j].to!=T) mat[e[j].to][i]=e[j].c;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                printf("%d",mat[i][j]);
                if(j==n) printf("\n");
                else printf(" ");
            }
        }
    }
}flow;
int main()
{
    cin.tie(0);
    cout.tie(0);
    cin>>n>>m;
    int sum=0,ss=0;
    flow.init(0,2*n+1);
    FOR(i,1,n) si(a[i]),ss+=a[i];
    FOR(i,1,n) si(b[i]),sum+=b[i];
    if(ss!=sum) return puts("NO"),0;
    FOR(i,1,n) flow.Add(0,i,a[i]),flow.Add(i+n,2*n+1,b[i]),flow.Add(i,i+n,inf);
    while(m--)
    {
        si(x),si(y);
        flow.Add(x,y+n,inf);
        flow.Add(y,x+n,inf);
    }
    if(flow.slove()!=sum) puts("NO");
    else puts("YES"),flow.print(n);
    return 0;
}