《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 >= 1 && old_color == canva[i - 1][j]) {
        floodFill(i - 1, j, n, m, new_color, canva);
    }
    if (i <= n - 2 && old_color == canva[i + 1][j]) {
        floodFill(i + 1, j, n, m, new_color, canva);
    }
    if (j >= 1 && old_color == canva[i][j - 1]) {
        floodFill(i, j - 1, n, m, new_color, canva);
    }
    if (j <= m - 2 && old_color == canva[i][j + 1]) {
        floodFill(i, j + 1, 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 = 0; i < n; ++i) {
        canva[i].resize(m);
    }
    
    for (i = 0; i < n; ++i) {
        for (j = 0; j < m; ++j) {
            scanf("%d", &canva[i][j]);
        }
    }
    
    while (scanf("%d%d%d", &i, &j, &c) == 3) {
        if (i >= 0 && i <= n - 1 && j >= 0 && j <= m - 1) {
            floodFill(i, j, n, m, c, canva);
        }
        for (i = 0; i < n; ++i) {
            for (j = 0; j < m; ++j) {
                printf((j == 0 ? "%d" : " %d"), canva[i][j]);
            }
            printf("
");
        }
    }
    
    for (i = 0; i < n; ++i) {
        canva[i].clear();
    }
    canva.clear();
    
    return 0;    
}