POJ3278 Catch That Cow
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本来以为是一道很简单的题 然后后来RE了10次
看了看讨论区才知道是数组越界问题
就是一定要先判断数组有没有越界
附AC BFS写法
#include<iostream>#include<queue>using namespace std;#define MAX 1000001#define INF 1000000void bfs(int n, int k);int d[MAX];bool sta[MAX];int main(){ int n, k; cin >> n >> k; { memset(d, 0, sizeof(d)); memset(sta, false, sizeof(sta)); bfs(n, k); cout << d[k] << endl; }}void bfs(int n, int k){ queue<int>que; que.push(n); while (que.size()) { int T = que.front(); int a, b, c; a = T-1; b =T+1; if(n<k) c = T * 2; if (n >= k) { d[k] = n - k; return; } que.pop(); if (a >= 0&&a<=MAX-5 && sta[a] == false) { que.push(a); d[a] = d[T] + 1; sta[a] = true; } if (b <= MAX - 5&&sta[b]==false) { que.push(b); d[b] = d[T] + 1; sta[b] = true; } if (c <= MAX - 5&&sta[c]==false) { que.push(c); d[c] = d[T] + 1; sta[c] = true; } if (sta[k]==true) return; }}
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