Game with Pearls(二分図)

Game with Pearls
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 145    Accepted Submission(s): 96
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
 
     タイトル:
     T組のサンプルを与え,NとKを後に与え,N本の管を表し,後にN本の管の上の初期状態の真珠数を与えた.いずれかのチューブにKの真珠数を加えて、チューブにそれぞれ1~Nの数を出すことができるかを聞くことができます.
 
     考え方:
     二分図、Nの最大値は100なので、図を建てるのもとても便利です.各数ansにKを加えてans>Nにループして図を作成し,最後に二分整合後に満配であるか否かを判断すればよい.
 
     AC:

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAX = 105;

int n, k;
int num[MAX];

int G[MAX][MAX];
int linker[MAX];
bool vis[MAX];

void build () {
    memset(G, 0, sizeof(G));

    for (int i = 1; i <= n; ++i) {
        for (int j = num[i]; j <= n; j += k) {
            G[i][j] = 1;
        }
    }
}

bool dfs (int x) {
    for (int i = 1; i <= n; ++i) {
        if (G[x][i] && !vis[i]) {
            vis[i] = 1;
            if (linker[i] == -1 || dfs(linker[i])) {
                linker[i] = x;
                return true;
            }
        }
    }

    return false;
}

int hungary () {
    int res = 0;
    memset(linker, -1, sizeof(linker));

    for (int i = 1; i <= n; ++i) {
        memset(vis, 0, sizeof(vis));
        if (dfs(i)) ++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", &num[i]);
        }

        build();

        if (n == hungary()) printf("Jerry
");
        else printf("Tom
");

    }

    return 0;
}