POJ - 3278 Catch That Cow 【BFS】

题目链接

http://poj.org/problem?id=3278

题意

给出两个数字 N K 每次 都可以用三个操作 + 1 - 1 * 2

求 最少的操作次数 使得 N 变成 K

思路

BFS 但是要注意 设置 数组的范围 小心 RE

AC代码

#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <list>
#include <numeric>
#include <sstream>
#include <iomanip>
#include <limits>

#define CLR(a, b) memset(a, (b), sizeof(a))
#define pb push_back

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair<string, int> psi;
typedef pair<string, string> pss;

const double PI = acos(-1.0);
const double E = exp(1.0);
const double eps = 1e-8;

const int INF = 0x3f3f3f3f;
const int maxn = 2e5 + 5;
const int MOD = 1e9 + 7;

int n, k;

int ans;

int v[maxn];

struct node
{
    int v, step;
};

bool ok(int x)
{
    if (x < 0 || x > maxn)
        return false;
    return true;
}

void bfs()
{
    CLR(v, 0);
    queue <node> q;
    node tmp;
    tmp.v = n;
    tmp.step = 0;
    q.push(tmp);
    v[n] = 1;
    while (!q.empty())
    {
        node u = q.front(), V;
        q.pop();
        if (u.v == k)
        {
            ans = u.step;
            return;
        }
        V.step = u.step + 1;
        V.v = u.v + 1;
        if (ok(V.v) && v[V.v] == 0)
        {
            q.push(V);
            v[V.v] = 1;
        }
        V.v = u.v - 1;
        if (ok(V.v) && v[V.v] == 0)
        {
            q.push(V);
            v[V.v] = 1;
        }
        V.v = u.v * 2;
        if (ok(V.v) && v[V.v] == 0)
        {
            q.push(V);
            v[V.v] = 1;
        }
    }
}

int main()
{
    scanf("%d%d", &n, &k);
    ans = INF;
    bfs();
    cout << ans << endl;
}