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

本是古典何须时尚 2022-08-05 00:43 24阅读 0赞

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 *a**i* 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 *b**i* 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 *a*1, *a*2, ..., *a**n* (0 ≤ *a**i* ≤ 100).

Next line contains *n* integers *b*1, *b*2, ..., *b**n* (0 ≤ *b**i* ≤ 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

简单网络流：

0---a\[u\]---> u ---INF---> v ---b\[v\]---> 2\*n+1

RE多次...

//      whn6325689
    //		Mr.Phoebe
    //		http://blog.csdn.net/u013007900
    #include <algorithm>
    #include <iostream>
    #include <iomanip>
    #include <cstring>
    #include <climits>
    #include <complex>
    #include <fstream>
    #include <cassert>
    #include <cstdio>
    #include <bitset>
    #include <vector>
    #include <deque>
    #include <queue>
    #include <stack>
    #include <ctime>
    #include <set>
    #include <map>
    #include <cmath>
    #include <functional>
    #include <numeric>
    #pragma comment(linker, "/STACK:1024000000,1024000000")
    
    
    using namespace std;
    
    typedef long long ll;
    typedef long double ld;
    typedef pair<ll, ll> pll;
    typedef complex<ld> point;
    typedef pair<int, int> pii;
    typedef pair<pii, int> piii;
    typedef vector<int> vi;
    
    #define CLR(x,y) memset(x,y,sizeof(x))
    #define mp(x,y) make_pair(x,y)
    #define pb(x) push_back(x)
    #define lowbit(x) (x&(-x))
    #define MID(x,y) (x+((y-x)>>1))
    #define speed std::ios::sync_with_stdio(false);
    #define eps 1e-9
    #define PI acos(-1.0)
    #define INF 0x3f3f3f3f
    #define LLINF 1LL<<62
    
    template<class T>
    inline bool read(T &n)
    {
        T x = 0, tmp = 1;
        char c = getchar();
        while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
        if(c == EOF) return false;
        if(c == '-') c = getchar(), tmp = -1;
        while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
        n = x*tmp;
        return true;
    }
    template <class T>
    inline void write(T n)
    {
        if(n < 0)
        {
            putchar('-');
            n = -n;
        }
        int len = 0,data[20];
        while(n)
        {
            data[len++] = n%10;
            n /= 10;
        }
        if(!len) data[len++] = 0;
        while(len--) putchar(data[len]+48);
    }
    //-----------------------------------
    
    const int M=4*210;
    const int N=4*111;
    
    struct ISAP
    {
        struct nedge
        {
            int u,v,next;
            ll c;
        } e[M*2];
        int head[N],tot,n;
        int ss,tt;
    
        void Resize(int n)
        {
            this->n=n;
        }
    
        void Clearall()
        {
            tot=1;
            CLR(head,0);
        }
    
        void Addedge(int u,int v,ll w)
        {
            e[++tot].v=v;
            e[tot].u=u;
            e[tot].c=w;
            e[tot].next=head[u];
            head[u]=tot;
            e[++tot].v=u;
            e[tot].u=v;
            e[tot].c=0;
            e[tot].next=head[v];
            head[v]=tot;
        }
    
        int dep[N],gap[N];
    
        void bfs()
        {
            CLR(dep,-1);
            CLR(gap,0);
            queue<int> que;
            gap[0]=1;
            dep[tt]=0;
            que.push(tt);
            while(!que.empty())
            {
                int u=que.front();
                que.pop();
                for(int i=head[u]; i; i=e[i].next)
                    if(dep[e[i].v]==-1)
                    {
                        que.push(e[i].v);
                        dep[e[i].v]=dep[u]+1;
                        ++gap[dep[e[i].v]];
                    }
            }
        }
    
        int res,cur[N],num[N];
        int top;
        int MaxFlow(int s,int t)
        {
            this->ss=s;
            this->tt=t;
            bfs();
            top=res=0;
            memcpy(cur,head,sizeof(head));
            int u=ss,i;
            while(dep[ss]<n)
            {
                if(u==tt)
                {
                    int temp=INF;
                    int inser;
                    for(i=0; i<top; i++)
                        if(temp>e[num[i]].c)
                        {
                            temp=e[num[i]].c;
                            inser=i;
                        }
                    for(i=0; i<top; i++)
                    {
                        e[num[i]].c-=temp;
                        e[num[i]^1].c+=temp;
                    }
                    res+=temp;
                    top=inser;
                    u=e[num[top]].u;
                }
                if(u!=tt && gap[dep[u]-1]==0)
                    break;
                for(i=cur[u]; i; i=e[i].next)
                    if(e[i].c != 0 && dep[u]==dep[e[i].v]+1)
                        break;
                if(i)
                {
                    cur[u]=i;
                    num[top++]=i;
                    u=e[i].v;
                }
                else
                {
                    int mi=n;
                    for(i=head[u]; i; i=e[i].next)
                        if(e[i].c>0 && mi>dep[e[i].v])
                        {
                            mi=dep[e[i].v];
                            cur[u]=i;
                        }
                    --gap[dep[u]];
                    dep[u]=mi+1;
                    ++gap[dep[u]];
                    if(u!=ss)
                        u=e[num[--top]].u;
                }
            }
            return res;
        }
    
        void print()
        {
            puts("GRAPH:");
            for(int i=0; i<=n; i++)
            {
                if(!head[i])
                    continue;
                printf("%d ->",i);
                for(int j=head[i]; j; j=e[j].next)
                    printf("%d(%I64d) ",e[j].v,e[j].c);
                putchar('\n');
            }
        }
    }g;
    
    int ans[111][111];
    int a[111],b[111];
    
    
    int main()
    {
    	int n,m,s1=0,s2=0;
    	read(n),read(m);
    	g.Clearall();
    	g.Resize(2*n+2);
    	for(int i=1;i<=n;i++)
    	{
    		read(a[i]);s1+=a[i];	
    		g.Addedge(0,i,a[i]);
    	}
    	for(int i=1;i<=n;i++)
    	{
    		read(b[i]);s2+=b[i];
    		g.Addedge(i,i+n,INF); 
    		g.Addedge(i+n,2*n+1,b[i]);
    	}	
    	for(int i=0,u,v;i<m;i++)
    	{
    		read(u),read(v);
    		g.Addedge(u,n+v,INF);
    		g.Addedge(v,n+u,INF);
    	}
    	if(s1!=s2)
    	{
    		printf("NO\n");
    	}
    	else if(g.MaxFlow(0,2*n+1)!=s1)
    	{
    		printf("NO\n");
    	}
    	else
    	{
    		printf("YES\n");
    		for(int i=1;i<=n;i++)
    		{
    			for(int j=g.head[i];j;j=g.e[j].next)
    			{
    				if(g.e[j].v==0)	continue;
    	 			ans[i][g.e[j].v-n]=INF-g.e[j].c;
    			}
    		}
    		for(int i=1;i<=n;i++)
    			for(int j=1;j<=n;j++)
    				printf("%d%c",ans[i][j],j!=n?' ':'\n');
    	}
    	return 0;
    }