HDOJ 5090 Game with Pearls bipartite graph matching, hdoj5090
Simple Bipartite Graph Matching:
The number of edges at each position can be the number of connected edges.
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 DescriptionTom 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 ".
InputThe 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.
OutputFor 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
Source2014 Shanghai national invitational competition-reproduction of the questions (thanks to Shanghai University for providing the questions)
#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;}