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  • Black And White(DFS+剪枝)

    Black And White

    Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
    Total Submission(s): 3937    Accepted Submission(s): 1082
    Special Judge


    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 ci 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 ci (ci > 0), denoting the number of cells where the i-th color should be used.

    It’s guaranteed that c1 + c2 + · · · + cK = 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
     
     
    //题意:第一行一个 T ,然后3个整数 n,m,k 代表有个 n 行 m 列的矩阵,k 代表有k种颜色,然后是 k 种颜色分别有多少个
    要用这些颜色涂满矩阵,相邻颜色不能相同,问能否填出
     
    DFS+剪枝,剪枝就是,如果有一种颜色比剩下的格子+1的一半还多的话,就 return
     1 #include <iostream>
     2 #include <stdio.h>
     3 #include <string.h>
     4 #include <stdlib.h>
     5 #include <math.h>
     6 #include <algorithm>
     7 #include <map>
     8 #include <stack>
     9 #include <queue>
    10 #include <set>
    11 #include <vector>
    12 using namespace std;
    13 #define LL long long
    14 #define PI acos(-1.0)
    15 #define lowbit(x) (x&(-x))
    16 #define INF 0x7f7f7f7f      // 21 E
    17 #define MEM 0x7f            // memset 都变为 INF
    18 #define MOD 4999            //
    19 #define eps 1e-9            // 精度
    20 #define MX  10         // 数据范围
    21 
    22 int read() {    //输入外挂
    23     int res = 0, flag = 0;
    24     char ch;
    25     if((ch = getchar()) == '-') flag = 1;
    26     else if(ch >= '0' && ch <= '9') res = ch - '0';
    27     while((ch = getchar()) >= '0' && ch <= '9') res = res * 10 + (ch - '0');
    28     return flag ? -res : res;
    29 }
    30 // code... ...
    31 int n,m,k;
    32 int ok;
    33 int color[MX*MX];
    34 int num[MX][MX];
    35 
    36 void dfs(int x,int y)
    37 {
    38     if (x>n) ok=1;
    39     for (int i=1;i<=k;i++)
    40     {
    41         int remain = n*m-((x-1)*m+y-1)+1;
    42         if (color[i]>remain/2) return;
    43     }
    44     for (int i=1;i<=k;i++)
    45     {
    46         if (color[i]>0&&num[x-1][y]!=i&&num[x][y-1]!=i)
    47         {
    48             num[x][y]=i;
    49             color[i]--;
    50             if (y==m) dfs(x+1,1);
    51             else dfs(x,y+1);
    52             color[i]++;
    53             if (ok) return;
    54         }
    55     }
    56 }
    57 
    58 int main()
    59 {
    60     int T=read();
    61     for (int cnt=1;cnt<=T;cnt++)
    62     {
    63         n=read();m=read();k=read();
    64         for (int i=1;i<=k;i++)
    65             color[i]=read();
    66         ok = 0;
    67         memset(num,0,sizeof(num));
    68         dfs(1,1);
    69         printf("Case #%d:
    ",cnt);
    70         if (ok)
    71         {
    72             printf("YES
    ");
    73             for (int i=1;i<=n;i++)
    74             {
    75                 for (int j=1;j<m;j++)
    76                     printf("%d ",num[i][j]);
    77                 printf("%d
    ",num[i][m]);
    78             }
    79         }
    80         else printf("NO
    ");
    81     }
    82     return 0;
    83 }
    View Code
     
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  • 原文地址:https://www.cnblogs.com/haoabcd2010/p/6776225.html
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