POJ 3278 / hdu 2717 Catch That Cow (广搜）

POJ 3278

HDU 2717

广搜题，用一个数组标记就可以过，不标记的话会超内存。

另外，poj的数据要比hdu强一些，比如0 100，这种数据。不特判的话会RE。不过如果不特判，在poj上用C++提交也可以过，不太清楚为什么，难道C++对于负数的数组访问会优化？期待大神解答。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
int n, k;
int vis[100100];
struct node {int pos, num;
};
void bfs() {node x;x.pos = n;x.num = 0;queue<node> q;vis[x.pos] = 1;q.push(x);while(!q.empty()) {node t = q.front();q.pop();if(t.pos == k) {printf("%d\n", t.num);return;}node tt;tt.pos = t.pos + 1, tt.num = t.num + 1;if(tt.pos <= 100000 && !vis[tt.pos]) {vis[tt.pos] = 1;q.push(tt);}if(t.pos > 0) {    //排除小于零的情况 tt.pos = t.pos - 1, tt.num = t.num + 1;if(tt.pos <= 100000 && !vis[tt.pos]) {vis[tt.pos] = 1;q.push(tt);}tt.pos = t.pos << 1, tt.num = t.num + 1;if(tt.pos <= 100000 && !vis[tt.pos]) {vis[tt.pos] = 1;q.push(tt);}}}
}
int main() {while(~scanf("%d %d", &n, &k)) {memset(vis, 0,sizeof(vis));bfs();}return 0;
}

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