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HDU 4975 A simple Gaussian elimination problem.(网络最大流)

2024-07-19 00:23:17   224人阅读

http://acm.hdu.edu.cn/showproblem.php?pid=4975

A simple Gaussian elimination problem.
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 669    Accepted Submission(s): 222

Problem Description
Dragon is studying math. One day, he drew a table with several rows and columns, randomly wrote numbers on each elements of the table. Then he counted the sum of each row and column. Since he thought the map will be useless after he got the sums, he destroyed the table after that.

However Dragon‘s mom came back and found what he had done. She would give dragon a feast if Dragon could reconstruct the table, otherwise keep Dragon hungry. Dragon is so young and so simple so that the original numbers in the table are one-digit number (e.g. 0-9).

Could you help Dragon to do that?
 

Input
The first line of input contains only one integer, T(<=30), the number of test cases. Following T blocks, each block describes one test case.

There are three lines for each block. The first line contains two integers N(<=500) and M(<=500), showing the number of rows and columns.

The second line contains N integer show the sum of each row.

The third line contains M integer show the sum of each column.
 

Output
Each output should occupy one line. Each line should start with "Case #i: ", with i implying the case number. For each case, if we cannot get the original table, just output: "So naive!", else if we can reconstruct the table by more than one ways, you should output one line contains only: "So young!", otherwise (only one way to reconstruct the table) you should output: "So simple!".
 

Sample Input
3
1 1
5
5
2 2
0 10
0 10
2 2
2 2
2 2
 

Sample Output
Case #1: So simple!
Case #2: So naive!
Case #3: So young!
 

Source
2014 Multi-University Training Contest 10
 


出题人是个蛤粉，蛤蛤蛤蛤蛤蛤蛤蛤蛤。。。。

题意：

每个格子只能填0~9这10个整数，给出行和及列和，求是否有合法方案，如果有并判断唯一性。

分析：

看到这题就感觉熟悉啊，和之前某场多校的题目是一样的，当初还不会网络流呢，现在虽然还不会建图，但是套套模板还是没有问题的。

建立二分图，行为X部，列为Y部，每个X部的点向Y部连一条容量为9的边，增加源点S，S向X部的所有点连边，容量为行和，增加汇点，每个Y部的点向汇点连边，容量为列和，在该图中跑一边网络最大流，如果满流则有合法方案。然后在残留网络中找环(不要立即走反向弧)，如果有环则有多种。找环之前最好重新建图，这样能避免判断大量满流边。

#include <cstdio>
#include <algorithm>
#include <cstring>
#define LL long long
#define itn int
#define maxn 1007
#define maxm 2333333
#define INF 0x3f3f3f3f
using namespace std;


int a[maxn],b[maxn];
int  fir[maxn];
itn u[maxm],v[maxm],cap[maxm],flow[maxm],nex[maxm];
int e_max;
itn q[maxn<<2];

itn lv[maxn],iter[maxn];

void add_edge(int _u,int _v,int _w)
{

    int e=e_max++;
    u[e]=_u;v[e]=_v;cap[e]=_w;
    nex[e]=fir[u[e]];fir[u[e]]=e;
    e=e_max++;
    u[e]=_v;v[e]=_u;cap[e]=0;
    nex[e]=fir[u[e]];fir[u[e]]=e;
}

void dinic_bfs(itn s)
{
    int f,r;
    lv[s]=0;
    q[f=r=0]=s;

    while (f<=r)
    {
        int x=q[f++];
        for (int e=fir[x];~e;e=nex[e])
        {
            if (cap[e]>flow[e] && lv[v[e]]<0)
            {
                lv[v[e]]=lv[u[e]]+1;
                q[++r]=v[e];
            }
        }
    }
}

int dinic_dfs(int s,int t,int _f)
{
    if (s==t)   return _f;

    for (int &e=iter[s];~e;e=nex[e])
    {
        if (cap[e]>flow[e] && lv[s]<lv[v[e]])
        {
            int _d=dinic_dfs(v[e],t,min(cap[e]-flow[e],_f));
            if (_d>0)
            {
                flow[e]+=_d;
                flow[e^1]-=_d;
                return _d;
            }
        }
    }

    return 0;
}


itn max_flow(int s,int t)
{
    int total_flow=0;

    memset(flow,0,sizeof flow);
    for (;;)
    {
        memset(lv,-1,sizeof lv);
        dinic_bfs(s);

        if (lv[t]==-1)  break;

        memcpy(iter,fir,sizeof fir);

        itn _f=0;
        while ((_f=dinic_dfs(s,t,INF))>0)
            total_flow+=_f;

    }

    return total_flow;

}

int vis[maxn];

bool dfs(itn s,int iter)
{
    for (int e=fir[s];~e;e=nex[e])
    {
        if ((e^1)!=iter)
        {
            if (vis[v[e]]==-1)  return true;
            vis[v[e]]=-1;
            if (dfs(v[e],e))    return true;
            vis[v[e]]=0;
        }

    }

    return false;
}

int main()
{
    int n,m;
    itn T_T,cas=0;
    scanf("%d",&T_T);

    while(T_T--)
    {
        printf("Case #%d: ",++cas);
        scanf("%d%d",&n,&m);
        itn s=0,t=n+m+1;
        itn sr=0,sc=0;
        e_max=0;
        memset(fir,-1,sizeof fir);
        for (int i=1;i<=n;i++)
        {
            scanf("%d",a+i);
            add_edge(s,i,a[i]);
            sr+=a[i];
        }

        for (int i=1;i<=m;i++)
        {
            scanf("%d",b+i);
            add_edge(i+n,t,b[i]);
            sc+=b[i];

        }

        if (sr!=sc)
        {
            printf("So naive!\n");
            continue;
        }


        for (int i=1;i<=n;i++)
        {
            for (int j=1;j<=m;j++)
            {
                add_edge(i,j+n,9);
            }
        }

        int res=max_flow(s,t);
        if (res!=sr)
        {
            printf("So naive!\n");
            continue;
        }

        itn mm=e_max;
        e_max=0;
        memset(fir,-1,sizeof fir);
        for (int e=0;e<mm;e++)
        {
            if (cap[e]>flow[e])
            {
                u[e_max]=u[e];
                v[e_max]=v[e];
                nex[e_max]=fir[u[e_max]];
                fir[u[e_max]]=e_max++;
            }
        }

        memset(vis,0,sizeof vis);
        bool cir=false;
        for (int i=1;i<=n;i++)
        {
            if (vis[i]==0 && dfs(i,-1))
            {
                cir=true;
                break;
            }
        }
        if (cir)
        {
            printf("So young!\n");
        }
        else
        {
            printf("So simple!\n");
        }
    }


    return 0;
}


HDU 4975 A simple Gaussian elimination problem.(网络最大流)


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