uva 11082

Posted 2020-08-21 焰

题意：知道矩阵的前i行之和，和前j列之和（任意i和j都可以）。求这个矩阵。每个格子中的元素必须在1~20之间。矩阵大小上限20*20

 

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#define pb push_back
using namespace std;
const int INF=0x3f3f3f3f;
int ans[25][25],R,C,r[25],c[25],A[25],B[25];
struct EdmondsKarp
{
    struct Edge
    {
        int from,to,cap,flow;
    };
    vector<Edge>edges;
    vector<int>G[45];
    int p[45],a[45];
    void init()
    {
        for(int i=0;i<45;i++)G[i].clear();
        edges.clear();
        memset(p,0,sizeof(p));
    }
    void addedge(int from,int to,int cap)
    {
        edges.pb((Edge){from,to,cap,0});
        edges.pb((Edge){to,from,0,0});
        int m=edges.size();
        G[from].pb(m-2);
        G[to].pb(m-1);
    }
    int maxflow(int s,int t)
    {
        int flow=0;
        while(1)
        {
            memset(a,0,sizeof(a));
            queue<int>q;
            q.push(s);
            a[s]=INF;
            while(!q.empty())
            {
                int x=q.front();
                q.pop();
                for(int i=0;i<G[x].size();i++)
                {
                    Edge& e=edges[G[x][i]];
                    if(!a[e.to]&&e.cap>e.flow)
                    {
                        p[e.to]=G[x][i];
                        a[e.to]=min(a[x],e.cap-e.flow);
                        q.push(e.to);
                    }
                }
                if(a[t])break;
            }
            if(!a[t])break;
            for(int u=t;u!=s;u=edges[p[u]].from)
            {
                edges[p[u]].flow+=a[t];
                edges[p[u]^1].flow-=a[t];
            }
            flow+=a[t];
        }
        return flow;
    }
    void solve()
    {
        int temp=maxflow(0,R+C+1);
        for(int len=edges.size(),i=0;i<len;i+=2)
        {
            Edge& e=edges[i];
            if(e.from!=0&&e.to!=R+C+1)
                ans[e.from][e.to-R]=e.flow+1;
        }
        for(int i=1;i<=R;i++)
            for(int j=1;j<=C;j++)
                printf(j==C?"%d\n":"%d ",ans[i][j]);
    }
}EK;
int main()
{
    int T;
    scanf("%d",&T);
    bool flag=0;
    for(int kase=1;kase<=T;kase++)
    {
        scanf("%d%d",&R,&C);
        EK.init();
        for(int i=1;i<=R;i++)scanf("%d",&r[i]);
        for(int i=1;i<=R;i++)A[i]=r[i]-r[i-1];
        for(int i=1;i<=C;i++)scanf("%d",&c[i]);
        for(int i=1;i<=C;i++)B[i]=c[i]-c[i-1];
        for(int i=1;i<=R;i++)
            EK.addedge(0,i,A[i]-C);
        for(int i=1;i<=C;i++)
            EK.addedge(i+R,R+C+1,B[i]-R);
        for(int i=1;i<=R;i++)
            for(int j=1;j<=C;j++)
                EK.addedge(i,R+j,19);
        if(kase>1)putchar(‘\n‘);
        printf("Matrix %d\n",kase);
        EK.solve();
    }
    return 0;
}