hdu1010:Tempter of the Bone

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1010

转自：http://www.cnblogs.com/zhourongqing/archive/2012/04/28/2475684.html

深度优先搜索，用到了奇偶剪枝，第一次听说。。仍然是咸鱼一条

奇偶剪枝：

是数据结构的搜索中，剪枝的一种特殊小技巧。

现假设起点为(sx,sy)，终点为（ex,ey），给定t步恰好走到终点，

如图所示（“|”竖走，“—”横走，“+”转弯），易证abs(ex-sx)+abs(ey-sy)为此问题类中任意情况下，起点到终点的最短步数，记做step，此处step1=8；

如图，为一般情况下非最短路径的任意走法举例，step2=14；

step2-step1=6，偏移路径为6，偶数（易证）；

故，若t-[abs(ex-sx)+abs(ey-sy)]结果为非偶数（奇数），则无法在t步恰好到达；

返回，false；

反之亦反。

刚开始也是写了宽搜，结果wa，看题，提取关键字！

代码：

  1 #define _CRT_SECURE_NO_WARNINGS
  2 #include <map>
  3 #include <set>
  4 #include <cmath>
  5 #include <queue>
  6 #include <stack>
  7 #include <cstdio>
  8 #include <string>
  9 #include <vector>
 10 #include <cstring>
 11 #include <iostream>
 12 #include <algorithm>
 13 #include <functional>
 14 #include<string.h>
 15 using namespace std;
 16 #define rep(i,a,n) for (int i=a;i<=n;i++)
 17 #define per(i,a,n) for (int i=n;i>=a;i--)
 18 #define pb push_back
 19 #define mp make_pair
 20 #define all(x) (x).begin(),(x).end()
 21 #define fi first
 22 #define se second
 23 #define SZ(x) ((int)(x).size())
 24 typedef vector<int> VI;
 25 typedef long long ll;
 26 typedef pair<int, int> PII;
 27 const ll mod = 1000000007;
 28 // head
 29 const int MAX_N = 10;
 30
 31 int i_add[] = { 0,1,0,-1 };
 32 int j_add[] = { 1,0,-1,0 };
 33
 34 int n, m, t;
 35 int stai, staj, endi, endj;
 36 char maze[MAX_N][MAX_N];
 37 bool visited[MAX_N][MAX_N];
 38 bool res = false;
 39
 40 bool dfs(int i, int j, int d) {
 41     if (res) return true;
 42     if (i < 0 || j < 0 || i >= n || j >= m) return false;
 43     if (d > t) return false;
 44     if (maze[i][j] == ‘D‘&&d == t) return res = true;
 45     int tmp = t - d - abs(i - endi) - abs(j - endj);
 46     if (tmp & 1) return false;
 47
 48     int ii, jj;
 49     ii = i - 1, jj = j;
 50     if (!visited[ii][jj] && maze[ii][jj] != ‘X‘) {
 51         visited[ii][jj] = true;
 52         dfs(ii, jj, d + 1);
 53         visited[ii][jj] = false;
 54     }
 55     ii = i, jj = j - 1;
 56     if (!visited[ii][jj] && maze[ii][jj] != ‘X‘) {
 57         visited[ii][jj] = true;
 58         dfs(ii, jj, d + 1);
 59         visited[ii][jj] = false;
 60     }
 61     ii = i + 1, jj = j;
 62     if (!visited[ii][jj] && maze[ii][jj] != ‘X‘) {
 63         visited[ii][jj] = true;
 64         dfs(ii, jj, d + 1);
 65         visited[ii][jj] = false;
 66     }
 67     ii = i, jj = j + 1;
 68     if (!visited[ii][jj] && maze[ii][jj] != ‘X‘) {
 69         visited[ii][jj] = true;
 70         dfs(ii, jj, d + 1);
 71         visited[ii][jj] = false;
 72     }
 73     return false;
 74 }
 75
 76 int main(){
 77     while (scanf("%d %d %d", &n, &m, &t)) {
 78         if (n == 0 && m == 0 && t == 0) {
 79             break;
 80         }
 81         memset(maze, 0, sizeof(maze));
 82         //initialize
 83         memset(visited, 0, sizeof(visited));
 84         int k = 0;
 85
 86         for (int i = 0; i < n; i++) {
 87             for (int j = 0; j < m; j++) {
 88                 scanf(" %c", &maze[i][j]);
 89                 if (maze[i][j] == ‘S‘) {
 90                     stai = i, staj = j;
 91                 }
 92                 if (maze[i][j] == ‘D‘) {
 93                     endi = i, endj = j;
 94                 }
 95                 if (maze[i][j] == ‘X‘) k++;
 96             }
 97         }
 98         if (n*m - k - 1 < t)
 99         {
100             cout << "NO" << endl;
101         }
102         else {
103             dfs(stai, staj, 0);
104             if (res)
105                 cout << "YES" << endl;
106             else
107                 cout << "NO" << endl;
108         }
109     }
110     return 0;
111 }