Hdoj 5090 Game with pearls binary map matching

Simple two-part graph matching:

Each position can be edged to these numbers and can even be

Game with PearlsTime 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 is 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 is several pearls. The number of pearls in each tube are at least 1 and at most N.

3) Jerry puts some more pearls to each tube. The number of pearls put into each tube have to be either 0 or a positive multiple of K. After this Jerry organizes these tubes in the order that the first tube have exact one pearl, the 2nd tube has exact 2 pear LS, ..., 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 a integer m (m<=500), then M. Games follow. For each game, the first line contains 2 integers, N and K (1 <= n <=, 1 <= K <= N), and the second line C Ontains N integers presenting the number of pearls in each tube.
Outputfor each game, output a line containing either "Tom" or "Jerry".
Sample Input

Sample Output

SOURCE2014 Shanghai National Invitational Competition--re-title (thanks to Shanghai University for its topics)

#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;}

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Hdoj 5090 Game with pearls binary map matching