hdu5113Black And White(深搜剪枝)

题目描述:

Black And White

Color an N × M chessboard with K colors numbered from 1 to K such that no two adjacent cells have the same color (two cells are adjacent if they share an edge). The i-th color should be used in exactly c_i cells.
Matt hopes you can tell him a possible coloring.

 

题目大意:给我们一个矩阵,再给出所有颜色的数目(总和等于矩阵大小),我们需要为所有空格涂上颜色,使得所有相邻空格颜色不相同,

思路:搜索与剪枝,先把所有颜色由大到小拍个序,即先选大的,可以更快找到可行解,这是一种优化思路,但是还是会超时,还有一个剪枝

就是判断(剩下的空格数+1)/2与是否能放下当前剩余个数最多的颜色,不能的话说明怎么放都放不好了,直接剪掉,顺利通关

AC代码:

#include<bits/stdc++.h>
using namespace std;
const int maxn = 50;
int m, n, k;
struct node {
    int num, id;
    bool operator<(node b) {
        return num < b.num;
    }
}a[maxn];
int maze[maxn][maxn];
bool dfs(int x, int y,int last) {
    if (last == 0)return true;
    for (int i = 1; i <= k; i++) {//如果剩下的空格数目的1/2比任何一个颜色个数要小,就放不下了,剪枝
        if ((last + 1) / 2 < a[i].num)return false;
    }
    for (int i = 1; i <= k; i++) {
        if (a[i].num == 0)continue;
        if (maze[x - 1][y] == a[i].id || maze[x][y - 1] == a[i].id)continue;
        a[i].num--; maze[x][y] = a[i].id;
        int tx = x, ty = y + 1;
        if (ty > n) {
            tx++; ty = 1;
        }
        if (dfs(tx, ty, last - 1))return true;
        a[i].num++; maze[x][y] = 0;
    }
    return false;
}
int main() {
    //freopen("test.txt", "r", stdin);
    int t; scanf("%d", &t);
    for (int cse = 1; cse <= t;cse++) {
        memset(maze, 0, sizeof(maze));
        scanf("%d%d%d", &m, &n, &k);
        for (int i = 1; i <= k; i++) {
            scanf("%d", &a[i].num);
            a[i].id = i;
        }
        sort(a + 1, a + k + 1);
        printf("Case #%d:\n", cse);
        if (dfs(1, 1, n * m)) {
            printf("YES\n");
            for (int i = 1; i <= m; i++) {
                for (int j = 1; j <= n; j++) {
                    printf("%d", maze[i][j]);
                        if (j != n)printf(" ");
                        else printf("\n");
                }
            }
        }
        else {
            printf("NO\n");
        }
    }
    return 0;
}