span att ould dash sam ssi fyi 上下 當前

題目鏈接

Problem Description 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 are required to color the regions of the map so that no two adjacent regions have the same color.
— Wikipedia, the free encyclopedia

In this problem, you have to solve the 4-color problem. Hey, I’m just joking.

You are asked to solve a similar problem:

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.

Input The first line contains only one integer T (1 ≤ T ≤ 5000), which indicates the number of test cases.

For each test case, the first line contains three integers: N, M, K (0 < N, M ≤ 5, 0 < K ≤ N × M ).

The second line contains K integers c_i (c_i > 0), denoting the number of cells where the i-th color should be used.

It’s guaranteed that c₁ + c₂ + · · · + c_K = N × M .

Output For each test case, the first line contains “Case #x:”, where x is the case number (starting from 1).

In the second line, output “NO” if there is no coloring satisfying the requirements. Otherwise, output “YES” in one line. Each of the following N lines contains M numbers seperated by single whitespace, denoting the color of the cells.

If there are multiple solutions, output any of them.

Sample Input 4 1 5 2 4 1 3 3 4 1 2 2 4 2 3 3 2 2 2 3 2 3 2 2 2

Sample Output Case #1: NO Case #2: YES 4 3 4 2 1 2 4 3 4 Case #3: YES 1 2 3 2 3 1 Case #4: YES 1 2 2 3 3 1

題意：有一個N*M的方格板，現在要在上面的每個方格上塗顏色，有K種顏色，每種顏色分別塗c[1]次、c[2]次……c[K]次，c[1]+c[2]+……+c[K]=N*M

要求每個方格的顏色與其上下左右均不同，如果可以輸出YES，並且輸出其中的一種塗法，如果不行，輸出NO；

思路：暴力搜索，但是這樣會超時，可以在搜索中加入剪枝：對於剩余的方格數res，以及當前剩余的顏色可塗數必須滿足（res+1）/2>=c[i]

否則在當前情況下繼續向下搜得不到正確塗法；

代碼如下：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
int N,K,M;
int c[30];
int mp[10][10];

int check(int x,int y,int k)
{
    int f=1;
    if(mp[x-1][y]==k) f=0;
    if(mp[x][y-1]==k) f=0;
    return f;
}

int cal(int x,int y)
{
    if(x>N) return 1;
    int res=(N-x)*M+M-y+2; ///剩余方格數+1 ；
    for(int i=1;i<=K;i++) if(res/2<c[i]) return 0; ///剪枝，某種顏色剩余方格數>(剩余方格數+1)/2 肯定不對；
    for(int i=1;i<=K;i++)
    {
        int f=0;
        if(c[i]&&check(x,y,i)){
            mp[x][y]=i; c[i]--;
            if(y==M)  f=cal(x+1,1);
            else      f=cal(x,y+1);
            c[i]++;
        }
        if(f) return 1;
    }
    return 0;
}

int main()
{
    int T,Case=1;
    cin>>T;
    while(T--)
    {
       scanf("%d%d%d",&N,&M,&K);
       for(int i=1;i<=K;i++) scanf("%d",&c[i]);
       printf("Case #%d:\n",Case++);

       if(!cal(1,1)) { puts("NO"); continue; }
       puts("YES");
       for(int i=1;i<=N;i++)
       for(int j=1;j<=M;j++)
         printf("%d%c",mp[i][j],(j==M)?‘\n‘:‘ ‘);
    }
    return 0;
}

HDU 5113--Black And White(搜索+剪枝)