HDOJ 5090 Game with Pearls 二分图匹配

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)
Total Submission(s): 122    Accepted Submission(s): 85

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;}