简单的二分图匹配：

每一个位置的数可能边成那些数连边即可

Game with Pearls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

Tom and Jerry are playing a game with tubes and pearls. The rule of the game is:

1) Tom and Jerry come up together with a number K.

2) Tom provides N tubes. Within each tube, there are several pearls. The number of pearls in each tube is at least 1 and at most N.

3) Jerry puts some more pearls into each tube. The number of pearls put into each tube has to be either 0 or a positive multiple of K. After that Jerry organizes these tubes in the order that the first tube has exact one pearl, the 2nd tube has exact 2 pearls,
…, the Nth tube has exact N pearls.

4) If Jerry succeeds, he wins the game, otherwise Tom wins.

Write a program to determine who wins the game according to a given N, K and initial number of pearls in each tube. If Tom wins the game, output “Tom”, otherwise, output “Jerry”.

Input

The first line contains an integer M (M<=500), then M games follow. For each game, the first line contains 2 integers, N and K (1 <= N <= 100, 1 <= K <= N), and the second line contains N integers presenting the number of pearls in each tube.

Output

For each game, output a line containing either “Tom” or “Jerry”.

Sample Input

2
5 1
1 2 3 4 5
6 2
1 2 3 4 5 5 

Sample Output

Jerry
Tom 

Source

2014上海全国邀请赛——题目重现（感谢上海大学提供题目）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=110;

int n,K;

struct Edge
{
    int to,next;
}edge[maxn*maxn];

int Adj[maxn],Size;

void init()
{
    memset(Adj,-1,sizeof(Adj)); Size=0;
}

void add_edge(int u,int v)
{
    edge[Size].to=v; edge[Size].next=Adj[u]; Adj[u]=Size++;
}

int linker[maxn];
bool used[maxn];

bool dfs(int u)
{
    for(int i=Adj[u];~i;i=edge[i].next)
    {
        int v=edge[i].to;
        if(!used[v])
        {
            used[v]=true;
            if(linker[v]==-1||dfs(linker[v]))
            {
                linker[v]=u;
                return true;
            }
        }
    }
    return false;
}

int hungary()
{
    int res=0;
    memset(linker,-1,sizeof(linker));
    for(int u=1;u<=n;u++)
    {
        memset(used,false,sizeof(used));
        if(dfs(u)) res++;
    }
    return res;
}

int a[maxn];

int main()
{
    int T_T;
    scanf("%d",&T_T);
    while(T_T--)
    {
        scanf("%d%d",&n,&K);
        init();
        for(int i=1;i<=n;i++)
        {
            scanf("%d",a+i);
            for(int j=0;a[i]+j*K<=n;j++)
            {
                int v=a[i]+j*K;
                add_edge(i,v);
            }
        }
        int pp=hungary();
        //cout<<"pp: "<<pp<<endl;
        if(pp==n) puts("Jerry");
        else puts("Tom");
    }
    return 0;
}