【九度OJ】题目1445：How Many Tables 解题报告

【九度OJ】题目1445：How Many Tables 解题报告

标签（空格分隔）： 九度OJ

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原题地址：http://ac.jobdu.com/problem.php?pid=1445

题目描述：

Today is Ignatius’ birthday. He invites a lot of friends. Now it’s dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all the friends do not want to stay with strangers.
One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, C know each other, so they can stay in one table.
For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and D, E have to stay in the other one. So Ignatius needs 2 tables at least.

输入：

The input starts with an integer T(1<=T<=25) which indicate the number of test cases. Then T test cases follow. Each test case starts with two integers N and M(1<=N,M<=1000). N indicates the number of friends, the friends are marked from 1 to N. Then M lines follow. Each line consists of two integers A and B(A!=B), that means friend A and friend B know each other. There will be a blank line between two cases.

输出：

For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.

样例输入：

2
5 3
1 2
2 3
4 5

5 1
2 5

样例输出：

2
4

Ways

根据上题的分析，只要有几个Tree[x]值为-1，那么就有单独几棵树，所以按照这个题的意思，就要有多少个桌子。所以这题和上题那个连通图没有区别。

另外，这个题目中的输入是有空行的，在C++中不用处理，因为会自动忽略掉这种输入。

#include<stdio.h>
#include<math.h>

int Tree[1002];

int findRoot(int x) {
    if (Tree[x] == -1) {
        return x;
    } else {
        int temp = findRoot(Tree[x]);
        Tree[x] = temp;
        return temp;
    }
}

int main() {
    int t;
    scanf("%d", &t);
    while (t-- != 0) {
        int n, m;
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; i++) {
            Tree[i] = -1;
        }
        while (m-- != 0) {
            int a, b;
            scanf("%d%d", &a, &b);
            int aRoot = findRoot(a);
            int bRoot = findRoot(b);
            if (aRoot != bRoot) {
                Tree[aRoot] = bRoot;
            }
        }
        int count = 0;
        for (int i = 1; i <= n; i++) {
            if (Tree[i] == -1) {
                count++;
            }
        }
        printf("%d\n", count);
    }

    return 0;
}

Date

2017 年 3 月 10 日