hdu 5090 二分匹配 or 排序

Game with Pearls

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 143 Accepted Submission(s): 95

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）  

二分匹配：当前数和当前数可以变成的数建边，求最大匹配

#include<stdio.h>#include<string.h>#define N 200int map[N][N],mark[N],pre[N],n,a[N];int find(int k){    int i;    for(i=1;i<=n;i++)    {        if(map[k][i]==1&&!mark[i])        {            mark[i]=1;            if(pre[i]==-1||find(pre[i]))            {                pre[i]=k;                return 1;            }        }    }    return 0;}int main(){    int t,b,i,k,sum;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&k);        memset(map,0,sizeof(map));        memset(pre,-1,sizeof(pre));        for(i=1;i<=n;i++)        {            scanf("%d",&a[i]);            b=a[i];            while(b<=n)            {                map[a[i]][b]=1;                b+=k;            }        }        sum=0;        for(i=1;i<=n;i++)        {            memset(mark,0,sizeof(mark));            sum+=find(a[i]);        }        if(sum==n)            printf("Jerry\n");        else            printf("Tom\n");    }    return 0;}

先排序，然后让当前没有匹配的最小的数加上k，然后判断是否合法，不合法的话，继续排序，继续让没有匹配的最小的数加上k，直到合法为止。

#include<stdio.h>#include<algorithm>#define N 200using namespace std;int a[N];int main(){    int t,n,k,i,j,flag;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&k);        for(i=1;i<=n;i++)            scanf("%d",&a[i]);        flag=0;        for(i=1;i<=n;i++)        {            sort(a+1,a+1+n);            for(j=1;j<=n;j++)            {                if(a[j]!=j)                {                    a[j]+=k;                    break;                }            }            if(j==n+1)            {                flag=1;                break;            }        }        if(flag==1)            printf("Jerry\n");        else            printf("Tom\n");    }    return 0;}