HDOJ 1010 Tempter of the Bone

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题意：一个N * M 的迷宫，起点为S，终点为D ，障碍为X, 问你是否恰好花费时间T 的时候到达终点D。
思路：DFS ，纯粹的搜索会直接超时，所以需要通过剪枝，也是在网上看到别人说奇偶剪枝，加进去直接AC了。

69234052012-10-15 15:09:01Accepted1010640MS256K1465BC++罗维

#include <iostream>#include <string>#include <vector>#include <math.h>using namespace std;struct point{int x, y;};int n, m, t;point spt, dpt;string maze[10];//vector<vecot<int> >used;bool escape;void dfs(point pt, int t){int tmp = t - abs(pt.x-dpt.x) - abs(pt.y - dpt.y);if(pt.x == dpt.x && pt.y == dpt.y && t == 0){escape = true;return ;}//剪枝if (escape || tmp < 0 || tmp&1 )  //tmp&1 如果是奇数则为1 否则为0return;maze[pt.x][pt.y] = 'X';if(pt.x + 1 <n && maze[pt.x + 1][pt.y] != 'X'){pt.x += 1;dfs(pt, t-1);pt.x -= 1;if(escape) return;}if(pt.x - 1 >=0 && maze[pt.x - 1][pt.y] != 'X'){pt.x -= 1;dfs(pt, t-1);pt.x += 1;if(escape) return;}if(pt.y + 1 < m && maze[pt.x][pt.y + 1] != 'X'){pt.y += 1;dfs(pt, t-1);pt.y -= 1;if(escape) return;}if(pt.y - 1 >= 0 && maze[pt.x][pt.y - 1] != 'X'){pt.y -= 1;dfs(pt, t-1);pt.y += 1;if(escape) return;}maze[pt.x][pt.y] = '.';}int main(){int i, j;while(cin>>n>>m>>t && n+m+t != 0){for (i=0; i<n; i++){cin>>maze[i];for (j=0; j<m; j++){if(maze[i][j] == 'S'){spt.x = i;spt.y = j;}else if(maze[i][j] == 'D'){dpt.x = i;dpt.y = j;}}}escape = false;dfs(spt, t);if(escape)cout<<"YES"<<endl;elsecout<<"NO"<<endl;}return 0;}