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.

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=205;//点数的最大值
const int MAXM=25005;//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=0;
    memset(head,-1,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=0;
    edge[tol].next=head[v];
    head[v]=tol++;
}
void BFS(int start,int end)
{
    memset(dep,-1,sizeof(dep));
    memset(gap,0,sizeof(gap));
    gap[0]=1;
    int que[MAXN];
    int front,rear;
    front=rear=0;
    dep[end]=0;
    que[rear++]=end;
    while(front!=rear)
    {
        int u=que[front++];
        if(front==MAXN)front=0;
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].to;
            if(dep[v]!=-1)continue;
            que[rear++]=v;
            if(rear==MAXN)rear=0;
            dep[v]=dep[u]+1;
            ++gap[dep[v]];
        }
    }
}
int SAP(int start,int end)
{
    int res=0;
    BFS(start,end);
    int cur[MAXN];
    int S[MAXN];
    int top=0;
    memcpy(cur,head,sizeof(head));
    int u=start;
    int i;
    while(dep[start]<n)
    {
        if(u==end)
        {
            int temp=INF;
            int inser;
            for(i=0;i<top;i++)
               if(temp>edge[S[i]].cap)
               {
                   temp=edge[S[i]].cap;
                   inser=i;
               }
            for(i=0;i<top;i++)
            {
                edge[S[i]].cap-=temp;
                edge[S[i]^1].cap+=temp;
            }
            res+=temp;
            top=inser;
            u=edge[S[top]].from;
        }
        if(u!=end&&gap[dep[u]-1]==0)//出现断层，无增广路
          break;
        for(i=cur[u];i!=-1;i=edge[i].next)
           if(edge[i].cap!=0&&dep[u]==dep[edge[i].to]+1)
             break;
        if(i!=-1)
        {
            cur[u]=i;
            S[top++]=i;
            u=edge[i].to;
        }
        else
        {
            int min=n;
            for(i=head[u];i!=-1;i=edge[i].next)
            {
                if(edge[i].cap==0)continue;
                if(min>dep[edge[i].to])
                {
                    min=dep[edge[i].to];
                    cur[u]=i;
                }
            }
            --gap[dep[u]];
            dep[u]=min+1;
            ++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 = 2 * n + 2;
        init();
        int s = n - 2,t = n - 1;
        sum = 0;
        for(int i = 0;i < N;i++)
        {
            scanf("%d",&sr);
            sum += sr;
            addedge(s,i,sr);
        }
        for(int i = 0;i < N;i++)
        {
            scanf("%d",&sc);
            addedge(N + i,t,sc);
        }
        for(int i = 0;i < N;i++)
        {
            for(int j = 0;j < N;j++)
            {
                addedge(i,N + j,100);
            }
        }
        if(SAP(s,t) == sum)
        {
            printf("YES
");
            int stk[105],sn = 0;
            for(int i = 0;i < N;i++)
            {
                for(int j = head[i];edge[j].next != -1;j = edge[j].next)
                {
                    stk[sn++] = 100 - edge[j].cap;
                }
                printf("%d",stk[--sn]);
                while(sn > 0)
                {
                    printf(" %d",stk[--sn]);
                }
                printf("
");
            }
        }
        else
            printf("NO
");
    }
    return 0;
}