HDU 5113 Black And White (dfs、剪枝)

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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5113

题意:N*M的棋盘,K种颜色,每种颜色有c[i]个(sigma(c[i]) = N*M),现在给棋盘染色,使得相邻的两个棋盘染成不同的颜色,并且把所有颜色用完。


就像数独一样,因为棋盘比较小,可以想到用搜索写。不过爆搜会超时,需要一个强有力的剪枝。

容易想到,当现在剩下的空格子数的一半是小于某一个c[i]的,那么这种颜色一定会在以后相邻,当然需要考虑一下剩下空格子的奇偶。加上了这个剪枝还是TLE.....

后来看了题解,发现思想和我想的一模一样。只不过人家对于空格子的奇偶判断的比我好,剪枝的界限比我小了1,然后就能过了。。。。时间还特快,62M。。。。

不太懂为什么,为什么就差这一点时间就会差这么多。

不过这题还是提醒我以后不要算着差不多就写,一定要搞精确了,不然改都不知道怎么改.......


#include <bits/stdc++.h>
using namespace std;
int n, m, k;
int grid[10][10];
int col[30];
int flag = 0;
void dfs(int x, int y, int blank) {
    //if(flag) return;
    if(x == n + 1) {
        flag = 1;
        return;
    }
    for(int i = 1; i <= k; i++) {
        if(col[i] > (blank + 1) / 2) return;  //剪枝的关键。 写成blank/2 + 1就会TLE,仔细分析一下确实(blank+1)/2更加精确一些。
    }
    int nx, ny;
    if(y == m) {
        nx = x + 1;ny = 1;
    }
    else {
        nx = x;ny = y + 1;
    }
    for(int i = 1; i <= k; i++) {
        if(col[i]) {
            if(x > 1 && grid[x - 1][y] == i) continue;
            if(y > 1 && grid[x][y - 1] == i) continue;
            col[i]--;
            grid[x][y] = i;
            dfs(nx, ny, blank - 1);
            if(flag) return;
            col[i]++;
        }
    }
}
int main() {
    int t;
    scanf("%d", &t);
    for(int kase = 1; kase <= t; kase++) {
        scanf("%d %d %d", &n, &m, &k);
        for(int i = 1; i <= k; i++) scanf("%d", col + i);
        flag = 0;
        memset(grid, 0, sizeof grid);
        dfs(1, 1, m * n);
        printf("Case #%d:\n", kase);
        if(flag) {
            puts("YES");
            for(int i = 1; i <= n; i++) {
                for(int j = 1; j <= m; j++) {
                    printf("%d", grid[i][j]);
                    if(j == m) puts("");
                    else putchar(' ');
                }
            }
        }
        else puts("NO");
    }
    return 0;
} 


用友YonSuite产品实操手把手教学(图文版) 用友YonSuite 阅读详情

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hdu5113 Black And White(dfs+剪枝)

题目链接:点击打开链接 题意描述:给一个n*m的棋盘,现在有k种颜色的涂料,每种涂料可以使用ai次。求能否找出一种方案,给所有的格子染色,保证相邻的格子之间颜色不同。如果存在给出任意一种解决方案;否则输出NO 解题思路:dfs+剪枝 由于题目中棋盘最大为5×5所以可以考虑使用dfs,染色问题有一个结论貌似是:(剩余的格子的数量+1)>= 任意一种涂料的个数,否则染色必然失败。因此我们

Crazy_Bear 634

hdu 5113 Black And White (dfs回溯+剪枝)

hdu 5113 Black And White (dfs回溯+剪枝)—— black 的专栏 —— waShaXiu

black的专栏 2293

hdu5113Black And White dfs暴搜 +剪枝

//n(n<=5)*m(m<=5)的格子 //k种颜料,每种颜料有ai个 //问这k种颜料能否将这n*m涂满且 //每相邻的格子颜色不同 //暴搜 , 剪枝剩下的颜色的数量大于剩下(sum+1)/2 //直接return #include #include #include using namespace std ; const int maxn = 30 ; int map[maxn][maxn

ijbuhv的专栏 640

DFS+剪枝 hdu 5113 Black And White

有一个包含从1到K一共K种颜色的N×M棋盘,使得任意两个相邻的区块不能有相同的颜色(如果它们的上、下、左、右任意一边的颜色与自身颜色不同)。第i种颜色可以被使用Ci次 思路:因为n,m较小,所以可以想到搜索,一行一行的给格子上色,首先这个颜色要有剩余并且不能和上面的和左边的颜色相同。 一个重要的剪枝:当剩下的格子个数为n时,任意一种颜色最多只能为(n+1)/2 #incl...

digu4869的博客 165

Black And White HDU - 5113 dfs+剪枝

In mathematics, the four color theorem, or the four color map theorem, states that, given any separation of a plane into contiguous regions, producing a figure called a map, no more than four colors a

mayuqing98的博客 222

HDU 5113 Black And White DFS+剪枝

In mathematics, the four color theorem, or the four color map theorem, states that, given any separation of a plane into contiguous regions, producing a figure called a map, no more than four colors a...

Harington的博客 280

hdu 5113 Black And White(dfs+剪枝)

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5113 解题思路: 题目大意: 给你一个N*M的棋盘,要求用K种颜色对每个格染色,相邻的格的颜色不能相同。并且第i种颜色必须恰好出现c[i]次。求是否存在该方案,如果存在,输出任意一种。 算法思想: 因为n,m都比较小,所以直接搜索即可,但是需要加一个剪枝,因为当剩下的容量为n时,

piaocoder 436

hdu 5113 Black And White dfs剪枝

这个。。。减枝方案看了题解才发现是自己思维太僵。。。 如果剩下的某种颜色大于剩下的格子数量的1/2那必有两个格子颜色相同然后return 鸽巢原理#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> using namespace std; int am[50],n,m,k,ans[10][10],f

acblacktea的博客 501
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