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2015
09-17

HDU 4707-Pet-DFS-[解题报告]HOJ

Pet

问题描述 :

One day, Lin Ji wake up in the morning and found that his pethamster escaped. He searched in the room but didn’t find the hamster. He tried to use some cheese to trap the hamster. He put the cheese trap in his room and waited for three days. Nothing but cockroaches was caught. He got the map of the school and foundthat there is no cyclic path and every location in the school can be reached from his room. The trap’s manual mention that the pet will always come back if it still in somewhere nearer than distance D. Your task is to help Lin Ji to find out how many possible locations the hamster may found given the map of the school. Assume that the hamster is still hiding in somewhere in the school and distance between each adjacent locations is always one distance unit.

输入:

The input contains multiple test cases. Thefirst line is a positive integer T (0<T<=10), the number of test cases. For each test cases, the first line has two positive integer N (0<N<=100000) and D(0<D<N), separated by a single space. N is the number of locations in the school and D is the affective distance of the trap. The following N-1lines descripts the map, each has two integer x and y(0<=x,y<N), separated by a single space, meaning that x and y is adjacent in the map. Lin Ji’s room is always at location 0.

输出:

The input contains multiple test cases. Thefirst line is a positive integer T (0<T<=10), the number of test cases. For each test cases, the first line has two positive integer N (0<N<=100000) and D(0<D<N), separated by a single space. N is the number of locations in the school and D is the affective distance of the trap. The following N-1lines descripts the map, each has two integer x and y(0<=x,y<N), separated by a single space, meaning that x and y is adjacent in the map. Lin Ji’s room is always at location 0.

样例输入:

1
10 2
0 1
0 2
0 3
1 4
1 5
2 6
3 7
4 8
6 9

样例输出:

2

题意:linji的仓鼠丢了,他要找回仓鼠,他在房间0放了一块奶酪,按照抓鼠手册所说,这块奶酪可以吸引距离它D的仓鼠,但是仓鼠还是没有出现,现在给出一张关系图,表示各个房间的关系,相邻房间距离为1,而且图中没有回路,每个房间都是联通的,求仓鼠可能出现的房间的数量。

很容易的dfs,50000个房间数据量比较大,用数组难以保存,于是用vector储存关系表。遍历过去,遍历过几个房间,那剩下的就是仓鼠可能出现的房间数了。

代码:

/*
*  Author:      illuz <iilluzen[at]gmail.com>
*  Blog:        http://blog.csdn.net/hcbbt
*  File:        2.cpp
*  Create Date: 2013-09-08 13:52:18
*  Descripton:  dfs 
*/

#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define mc(a) memset(a, 0, sizeof(a))

const int MAXN = 1e5 + 1;
vector<int> v[MAXN];
int n, d, cnt;
bool vis[MAXN];

void dfs(int p, int e) {
	if (e > d) return;
	vis[p] = 1;
	cnt++;
	int len = v[p].size();
	rep(i, len)
		if (!vis[v[p][i]])
			dfs(v[p][i], e + 1);
}

int main() {
	int t, a, b;
	scanf("%d", &t);

	while (t--) {
		rep(i, n) v[i].clear();
		mc(vis);
		cnt = 0;
		scanf("%d%d", &n, &d);
		rep(i, n - 1) {
		   	scanf("%d%d", &a, &b);
			v[a].push_back(b);
			v[b].push_back(a);
		}
		dfs(0, 0);
		printf("%d\n", n - cnt);
	}
	return 0;
}

版权声明:本文为博主原创文章,未经博主允许不得转载。

参考:http://blog.csdn.net/hcbbt/article/details/11401085


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