toj 1421

题意：假如存在矩阵A，A[i][0] + A[i][1] + ...... + A[i][n - 1] == SR[i],A[0][j] + A[1][j] + ...... + A[n - 1][j] == SC[j].(0 <= A[i][j] <= 100),输出YES并输出任意一个满足条件的矩阵A,若不存在则输出NO

分析:SR[0] + SR[1] + ...... + SR[n - 1] == SC[0] + SC[1] + ...... + SC[n - 1],建图的重点是把银行和公司之间的边的容量都设为100,作为流的上限.

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" alt="" />当图的最大流等于SR的和则找到了满足条件的矩阵.

另外,输出矩阵时,由于建图方法的特殊性,要用栈来倒序输出.

#include <cstdio>
#include <cstring>
const int MAXN=;//点数的最大值
const int MAXM=;//边数的最大值
const int INF=0x3fffffff;
 
struct Node
{
    int from,to,next;
    int cap;
}edge[MAXM];
 
int tol;
int head[MAXN];
int dep[MAXN];
int gap[MAXN];//gap[x]=y :说明残留网络中dep[i]==x的个数为y
 
int n;//n是总的点的个数，包括源点和汇点
 
void init()
{
    tol=;
    memset(head,-,sizeof(head));
}
 
void addedge(int u,int v,int w)
{
    edge[tol].from=u;
    edge[tol].to=v;
    edge[tol].cap=w;
    edge[tol].next=head[u];
    head[u]=tol++;
    edge[tol].from=v;
    edge[tol].to=u;
    edge[tol].cap=;
    edge[tol].next=head[v];
    head[v]=tol++;
}
void BFS(int start,int end)
{
    memset(dep,-,sizeof(dep));
    memset(gap,,sizeof(gap));
    gap[]=;
    int que[MAXN];
    int front,rear;
    front=rear=;
    dep[end]=;
    que[rear++]=end;
    while(front!=rear)
    {
        int u=que[front++];
        if(front==MAXN)front=;
        for(int i=head[u];i!=-;i=edge[i].next)
        {
            int v=edge[i].to;
            if(dep[v]!=-)continue;
            que[rear++]=v;
            if(rear==MAXN)rear=;
            dep[v]=dep[u]+;
            ++gap[dep[v]];
        }
    }
}
int SAP(int start,int end)
{
    int res=;
    BFS(start,end);
    int cur[MAXN];
    int S[MAXN];
    int top=;
    memcpy(cur,head,sizeof(head));
    int u=start;
    int i;
    while(dep[start]<n)
    {
        if(u==end)
        {
            int temp=INF;
            int inser;
            for(i=;i<top;i++)
               if(temp>edge[S[i]].cap)
               {
                   temp=edge[S[i]].cap;
                   inser=i;
               }
            for(i=;i<top;i++)
            {
                edge[S[i]].cap-=temp;
                edge[S[i]^].cap+=temp;
            }
            res+=temp;
            top=inser;
            u=edge[S[top]].from;
        }
        if(u!=end&&gap[dep[u]-]==)//出现断层，无增广路
          break;
        for(i=cur[u];i!=-;i=edge[i].next)
           if(edge[i].cap!=&&dep[u]==dep[edge[i].to]+)
             break;
        if(i!=-)
        {
            cur[u]=i;
            S[top++]=i;
            u=edge[i].to;
        }
        else
        {
            int min=n;
            for(i=head[u];i!=-;i=edge[i].next)
            {
                if(edge[i].cap==)continue;
                if(min>dep[edge[i].to])
                {
                    min=dep[edge[i].to];
                    cur[u]=i;
                }
            }
            --gap[dep[u]];
            dep[u]=min+;
            ++gap[dep[u]];
            if(u!=start)u=edge[S[--top]].from;
        }
    }
    return res;
}
int main()
{
    int sr,sc,sum;
    while(~scanf("%d",&n))
    {
        int N = n;
        n =  * n + ;
        init();
        int s = n - ,t = n - ;
        sum = ;
        for(int i = ;i < N;i++)
        {
            scanf("%d",&sr);
            sum += sr;
            addedge(s,i,sr);
        }
        for(int i = ;i < N;i++)
        {
            scanf("%d",&sc);
            addedge(N + i,t,sc);
        }
        for(int i = ;i < N;i++)
        {
            for(int j = ;j < N;j++)
            {
                addedge(i,N + j,);
            }
        }
        if(SAP(s,t) == sum)
        {
            printf("YES\n");
            int stk[],sn = ;
            for(int i = ;i < N;i++)
            {
                for(int j = head[i];edge[j].next != -;j = edge[j].next)
                {
                    stk[sn++] =  - edge[j].cap;
                }
                printf("%d",stk[--sn]);
                while(sn > )
                {
                    printf(" %d",stk[--sn]);
                }
                printf("\n");
            }
        }
        else
            printf("NO\n");
    }
    return ;
}