hdu5090 匈牙利算法二分图最大匹配问题

http://acm.hdu.edu.cn/showproblem.php?pid=5090

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

/**
hdu5090二分图的最大匹配
一开始我用的是遍历写的，知道必然是TLE还是试了试，赛后才知道是一个二分图的题，无奈整了这么久的图论竟然看不出这是二分图，真是愧对队友啊==
解题思路：因为最后的数是无序的，我们把每一个a[i]与二分图另一侧的a[i],a[i]+k……知道a[i]+n*k>n为止的所有点连一条边，
最后匈牙利一遍如果满流就可以实现，否则就呵呵了。
*/
/* **************************************************************************
//二分图匹配（匈牙利算法的DFS实现）
//初始化：g[][]两边顶点的划分情况
//建立g[i][j]表示i->j的有向边就可以了，是左边向右边的匹配
//g没有边相连则初始化为0
//uN是匹配左边的顶点数，vN是匹配右边的顶点数
//调用：res=hungary();输出最大匹配数
//优点：适用于稠密图，DFS找增广路，实现简洁易于理解
//时间复杂度:O(VE)
//***************************************************************************/
//顶点编号从0开始的
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;

const int MAXN=220;
int n,k;//u,v数目
int g[MAXN][MAXN];
int linker[MAXN],a[220];
bool used[MAXN];

bool dfs(int u)//从左边开始找增广路径
{
int v;
for(v=1; v<=n; v++) //这个顶点编号从0开始，若要从1开始需要修改
if(g[u][v]&&!used[v])
{
used[v]=true;
if(linker[v]==-1||dfs(linker[v]))
{
//找增广路，反向
linker[v]=u;
return true;
}
}
return false;//这个不要忘了，经常忘记这句
}

int hungary()
{
int res=0;
int u;
memset(linker,-1,sizeof(linker));
for(u=1; u<=n; u++)
{
memset(used,0,sizeof(used));
if(dfs(u)) res++;
}
return res;
}
//******************************************************************************/
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&k);
for(int i=1; i<=n; i++)
scanf("%d",&a[i]);
memset(g,0,sizeof(g));
for(int i=1; i<=n; i++)
{
for(int j=1; j<=n; j++)
{
if((j-a[i])%k==0&&j>=a[i])
g[i][j]=1;
}
}
if(hungary()==n)
printf("Jerry\n");
else
printf("Tom\n");
}
return 0;
}

遍历的写法虽然TLE还是放这里吧，费了我不少心血呢==

#include <stdio.h>
#include <string.h>
#include <iostream>'
using namespace std;
int n,k,sa[200],s[200][200],b[200],flag[200],cont;
void dfs(int x)
{
if(x==n+1)
{
int tt=0;
for(int i=1;i<=n;i++)
if(flag[i]==1)
tt++;
if(tt==n)
cont=1;
return;
}
for(int i=0;i<=b[x];i++)
{
if(cont==1)
return;
flag[s[x][i]]=1;
dfs(x+1);
flag[s[x][i]]=0;
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&k);
memset(b,0,sizeof(b));
for(int i=1;i<=n;i++)
{
scanf("%d",&sa[i]);
s[i][b[i]]=sa[i];
for(int j=1;j<=n;j++)
{
//printff("(%d %d)\n",b[i],s[i][b[i]]);
if(s[i][b[i]]+k>n)
break;
b[i]++;
s[i][b[i]]=s[i][b[i]-1]+k;
}
}
// for(int i=1;i<=n;i++)
//       printff("%d ",b[i]);
//     printff("\n");
/*for(int i=1;i<=n;i++)
{
for(int j=0;j<=b[i];j++)
printff("%d ",s[i][j]);
printff("\n");
}*/
cont=0;
memset(flag,0,sizeof(flag));
dfs(1);
if(cont==0)
printff("Tom\n");
else
printff("Jerry\n");
}
return 0;
}