Sunday, November 29, 2015

Least Jumps of Frog Cross River - 我的博客 - ITeye技术网站


Least Jumps of Frog Cross River - 我的博客 - ITeye技术网站
跳河问题。给一个0/1数组R代表一条河,0代表水,1代表石头。起始位置R[0]等于1,
初速度为1. 每一步可以选择以当前速度移动,或者当前速度加1再移动。只能停留在石
头上。问最少几步可以跳完整条河流。

给定数组为R=[1,1,1,0,1,1,0,0],最少3步能过河:
第一步先提速到2,再跳到R[2];
第二步先提速到3,再跳到R[5];
第三步保持速度3,跳出数组范围,成功过河。
http://www.mitbbs.com/article_t1/JobHunting/32617501_0_1.html
DP状态方程:f(i,j) = min( f(i-j, j), f(i-j, j-1) ) + 1
i: 当前石头在数组中所处的index (0 based)
j: 到达i时的可能最大步数, <= sqrt(i*2) 假设处处有石子,那么青蛙可以随意跳跃,每次跳跃步伐比前一次跳跃大1,就有1+2+..+j = j(j+1)/2 = i, j^2+j = i*2, 所以 j <= sqrt(i*2) 
f(i, j): 以步数j到达i时所需的最少跳跃次数,其中f(0,1)=0
时间和空间复杂度均为O(n^(3/2))
  1. public int minFrogJumps(int[] a) {  
  2.     int len = a.length;  
  3.     int[][] f = new int[len][(int)Math.sqrt(len * 2) + 1];  
  4.     int ans = Integer.MAX_VALUE;  
  5.     for (int i = 1; i < len; i++) {  
  6.         Arrays.fill(f[i], -1);  
  7.         if (a[i] == 0)  {  
  8.             continue;  
  9.         }  
  10.         for (int j = 1; j <= (int)Math.sqrt(i * 2); j++) {  
  11.             if (a[i - j] == 0) {  
  12.                 f[i][j] = -1;  
  13.                 continue;  
  14.             }  
  15.             if (f[i - j][j] == -1 && f[i - j][j - 1] == -1) {  
  16.                 continue;  
  17.             }  
  18.             if (f[i - j][j] != -1) f[i][j] = f[i - j][j] + 1;  
  19.             if (j > 1 && f[i - j][j - 1] != -1) {  
  20.                 if (f[i][j] == -1)  
  21.                     f[i][j] = f[i - j][j - 1] + 1;  
  22.                 else  
  23.                     f[i][j] = Math.min(f[i - j][j - 1] + 1, f[i][j]);  
  24.             }  
  25.             if (i + j + 1 >= len) {  
  26.                 ans = Math.min(f[i][j] + 1, ans);  
  27.             }  
  28.         }  
  29.     }  
  30.     return ans;  

  1. int FlogCrossRiver(string river) {  
  2.   if (river.empty()) {  
  3.     return 0;  
  4.   }  
  5.   vector<vector<pair<size_tint>>> vp(river.size());  
  6.   vp[0].emplace_back(1, 1);  
  7.   int res = INT_MAX;  
  8.   for (size_t i = 0; i < vp.size(); ++i) {  
  9.     if (river[i] == '0') {  
  10.       continue;  
  11.     }  
  12.     for (auto pr : vp[i]) {  
  13.       if (i + pr.first >= vp.size()) {  
  14.         res = min(pr.second, res);  
  15.       } else if (river[i + pr.first] == '1') {  
  16.         vp[i + pr.first].emplace_back(pr.first, pr.second + 1);  
  17.       }  
  18.       if (i + pr.first + 1 >= vp.size()) {  
  19.         res = min(pr.second, res);  
  20.       } else if (river[i + pr.first + 1] == '1') {  
  21.         vp[i + pr.first + 1].emplace_back(pr.first + 1, pr.second + 1);  
  22.       }  
  23.     }  
  24.   }  
  25.   return res;  
  26. }  
  27. void FlogCrossRiverTest() {  
  28.   cout << "11101100t" << FlogCrossRiver("11101100") << endl;  
  29.   cout << "11111111t" << FlogCrossRiver("11111111") << endl;  
  30.   cout << "11101000t" << FlogCrossRiver("11101000") << endl;  
  31.   cout << "11010101t" << FlogCrossRiver("11010101") << endl;  

Related: Minimum number of jumps to reach end
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