poj1741，POJ3278 HDU2717 Catch That Cow【BFS】

Catch That Cow

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.

* Walking: FJ can move from any point X to the points X - 1 orX + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

poj1741。Input

Line 1: Two space-separated integers: N andK

Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

Sample Input

5 17

poj2352、Sample Output

4

Hint

The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.

Source

USACO 2007 Open Silver

poj3273，?

問題鏈接：POJ3278 HDU2717 Catch That Cow

題意簡述：

? 一條線上，人的FJ的起點為K位置，牛在N位置（牛不動），輸入正整數K和N。若FJ在x位置，FJ有三種走法，分別是走到x-1、x+1或2x位置。求從K走到N的最少步數。

問題分析：

? 典型的BFS問題。在BFS搜索過程中，走過的點就不必再走了，因為這次再走下去不可能比上次的步數少。

程序說明：

? 程序中，使用了一個隊列來存放中間節點。

? 需要說明的是，除了BFS方法，這個題應該可以用分支限界法來解，需要更高的技巧。

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AC的C++語言程序如下：

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/* POJ3278 HDU2717 Catch That Cow */#include <iostream>
#include <cstring>
#include <queue>using namespace std;const int MAXN = 100000;
const int MAXN2 = MAXN * 2;bool notvist[MAXN * 2 + 2];int n, k, ans;struct node {int p, level;
};node start;void bfs()
{queue<node> q;int nextp;memset(notvist, true, sizeof(notvist));notvist[n] = false;start.p = n;start.level = 0;ans = 0;q.push(start);while(!q.empty()) {node front = q.front();q.pop();if(front.p == k) {ans = front.level;break;}nextp = front.p - 1;        /* x-1 */if(nextp >= 0 && notvist[nextp]) {notvist[nextp] = false;node v;v.p = nextp;v.level = front.level + 1;q.push(v);}nextp = front.p + 1;        /* x+1 */if(nextp <= MAXN2 && notvist[nextp]) {notvist[nextp] = false;node v;v.p = nextp;v.level = front.level + 1;q.push(v);}nextp = front.p + front.p;      /* 2x */if(nextp <= MAXN2 && notvist[nextp]) {notvist[nextp] = false;node v;v.p = nextp;v.level = front.level + 1;q.push(v);}}
}int main()
{while(cin >> n >> k) {bfs();printf("%d\n", ans);}return 0;
}

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