《Cracking the Coding Interview》——第9章：递归和动态规划——题目7

2014-03-20 03:35

题目：实现画图的Flood Fill操作。

解法：DFS和BFS皆可，但BFS使用的队列在时间复杂度上常数项比较大，速度略慢，所以我选了DFS。当然，如果图很大的话DFS是会导致call stack溢出的，那就摊上事儿了。

代码：

 // 9.7 Implement a flood fill painter that changes a certain area to a certain color. You are given one point as the seed.
 #include <cstdio>
 #include <vector>
 using namespace std;

 void floodFill(int i, int j, int n, int m, int new_color, vector<vector<int> > &canva)
 {
     int old_color = canva[i][j];

     canva[i][j] = new_color;
     if (i >=  && old_color == canva[i - ][j]) {
         floodFill(i - , j, n, m, new_color, canva);
     }
     if (i <= n -  && old_color == canva[i + ][j]) {
         floodFill(i + , j, n, m, new_color, canva);
     }
     if (j >=  && old_color == canva[i][j - ]) {
         floodFill(i, j - , n, m, new_color, canva);
     }
     if (j <= m -  && old_color == canva[i][j + ]) {
         floodFill(i, j + , n, m, new_color, canva);
     }
 }

 int main()
 {
     int i, j, c;
     int n, m;
     vector<vector<int> > canva;

     scanf("%d%d", &n, &m);
     canva.resize(n);
     for (i = ; i < n; ++i) {
         canva[i].resize(m);
     }

     for (i = ; i < n; ++i) {
         for (j = ; j < m; ++j) {
             scanf("%d", &canva[i][j]);
         }
     }

     while (scanf("%d%d%d", &i, &j, &c) == ) {
         if (i >=  && i <= n -  && j >=  && j <= m - ) {
             floodFill(i, j, n, m, c, canva);
         }
         for (i = ; i < n; ++i) {
             for (j = ; j < m; ++j) {
                 printf((j ==  ? "%d" : " %d"), canva[i][j]);
             }
             printf("\n");
         }
     }

     for (i = ; i < n; ++i) {
         canva[i].clear();
     }
     canva.clear();

     return ;
 }