poj 3278 catch that cow
2012-03-29 20:33
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bfs,RE多次之后学习大牛的代码,不要惊讶怎么限定在100000范围内!
更好的解法:http://www.cnblogs.com/longzhiri/articles/1555344.html
更好的解法:http://www.cnblogs.com/longzhiri/articles/1555344.html
# include <stdio.h> # include <string.h> # include <queue> using namespace std; int n, k, x; int d[100002]; queue<int>Q; int main() { while (~scanf("%d%d", &n, &k)) { memset(d, 0, sizeof(d)); while (!Q.empty()) Q.pop(); Q.push(n); while (!Q.empty()) { x = Q.front(); Q.pop(); if (x == k) break; if (x > 0 && 0==d[x-1]) {Q.push(x-1); d[x-1] = d[x]+1;} if (x < 100001 && 0==d[x+1]) {Q.push(x+1); d[x+1] = d[x]+1;} if ((x<<1) < 100001 && 0==d[x<<1]) {Q.push(x<<1); d[x<<1] = d[x]+1;} } printf("%d\n", d[k]); } return 0; }
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