原题: ZOJ 3675 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3675
由m<=20可知,可用一个二进制数表示指甲的状态,最多2^20,初始状态为0,表示指甲都没剪,然后BFS找解,每次枚举剪刀的两个方向,枚举移动的位数进行扩展状态即可。
代码:
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <queue> using namespace std; #define N 10007 struct node { int state,step; node(int _state,int _step) { state = _state; step = _step; } node(){} }; int vis[1<<21]; int cut[2]; //两个方向 queue<node> que; int n,m; int bfs(int s) { int i,j,k; memset(vis,0,sizeof(vis)); while(!que.empty()) que.pop(); int E = (1<<m)-1; que.push(node(s,0)); vis[s] = 1; while(!que.empty()) { node tmp = que.front(); que.pop(); int state = tmp.state; int step = tmp.step; int tms = state; for(i=0;i<2;i++) //direction { for(j=0;j<n;j++) //move { int end = ((cut[i]>>j) | tms) & E; // &E : keep m bit if(vis[end]) continue; vis[end] = 1; if(end == E) return step+1; que.push(node(end,step+1)); } for(j=0;j<m;j++) { int to = ((cut[i]<<j) | tms) & E; if(vis[to]) continue; vis[to] = 1; if(to == E) return step+1; que.push(node(to,step+1)); } } } return -1; } int main() { int i,j; char ss[13]; while(scanf("%d",&n)!=EOF) { cut[0] = cut[1] = 0; scanf("%s",ss); for(i=0;i<=n;i++) { if(ss[i] == ‘*‘) { cut[0] |= (1<<i); cut[1] |= (1<<(n-1-i)); } } scanf("%d",&m); if(cut[0] == 0) { puts("-1"); continue; } printf("%d\n",bfs(0)); } return 0; }
2014 Super Training #6 G Trim the Nails --状态压缩+BFS,布布扣,bubuko.com
时间: 2024-10-12 17:02:29