HDU 5113 Black And White（深搜+剪枝）

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

Black And White

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 4041    Accepted Submission(s): 1103
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 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

41 5 24 13 3 41 2 2 42 3 32 2 23 2 32 2 2

 

Sample Output

Case #1:NOCase #2:YES4 3 42 1 24 3 4Case #3:YES1 2 32 3 1Case #4:YES1 22 33 1

 

深搜+剪枝的题 只需要输出一种 

如果剩余没填的方格小于最多剩余颜色的两倍 就不能保证相邻的颜色不同 这是本题的关键

#include <stdio.h>#include <stdlib.h>#include <iostream>#include<algorithm>#include <math.h>#include <string.h>#include <limits.h>#include <string>#include <queue>#include <stack>using namespace std;int n=0;int m=0;int k=0;int num[26];int flag=0;int a[6][6];void dfs(int x,int y,int sy){        if(x<=0||x>n||y<=0||y>m||flag||sy<0)        return ;        for(int i=1;i<26;i++)    if((sy+1)<num[i]*2)        return;    if(sy==0)    {        flag=1;        printf("YES\n");        for(int i=1;i<=m;i++)                {                    printf("%d",a[1][i]);                    for(int j=2;j<=n;j++)                        printf(" %d",a[j][i]);                    printf("\n");                }        return ;    }    for(int i=1;i<=k;i++)    {        if(((a[x+1][y]!=i&&a[x-1][y]!=i&&a[x][y+1]!=i&&a[x][y-1]!=i)&&num[i]&&!a[x][y]))        {            num[i]--;            a[x][y]=i;            dfs(x+1,y,sy-1);            dfs(x-1,y,sy-1);            dfs(x,y+1,sy-1);            dfs(x,y-1,sy-1);            a[x][y]=0;            num[i]++;        }    }    return;}int main(){    int t=0;    int ca=1;    scanf("%d",&t);    while(t--)    {        for(int i=1;i<26;i++)            num[i]=0;        for(int i=1;i<=5;i++)            for(int j=1;j<=5;j++)            {                a[i][j]=0;            }        flag=0;        scanf("%d%d%d",&m,&n,&k);        for(int i=1;i<=k;i++)            scanf("%d",&num[i]);        printf("Case #%d:\n",ca);        if(m==1&&n==1&&k==1)        {            printf("YES\n");            printf("1\n");        }        else        {        dfs(1,1,n*m);        if(flag==0)            printf("NO\n");        }        ca++;            }}

又写了一遍。。格式问题wa到死 格式问题也不是pe

#include <bits/stdc++.h>using namespace std;int n,m,k;int cnt[30];int mmap[10][10];int flag=0;int book [10][10][30];int num=0;void dfs(int row,int clu){    if(row==n+1&&n!=0)    {        printf("YES\n");        for(int i=1;i<=n;i++)        {            printf("%d",mmap[i][1]);            for(int j=2;j<=m;j++)                printf(" %d",mmap[i][j]);            printf("\n");        }        flag=1;        return ;    }    for(int i=1;i<26;i++)    if((num+1)<cnt[i]*2)        return;    if(flag)        return ;    for(int j=1;j<=k;j++)    if(book[row][clu][j]==0&&cnt[j])        {            book[row][clu][j]+=1;            book[row+1][clu][j]+=1;            book[row][clu+1][j]+=1;            cnt[j]--;            mmap[row][clu]=j;            num--;            if(clu==m)                dfs(row+1,1);            else                dfs(row,clu+1);            num++;            cnt[j]++;            mmap[row][clu]=0;            book[row][clu][j]-=1;            book[row+1][clu][j]-=1;            book[row][clu+1][j]-=1;        }        return;}int main(){    int t;    scanf("%d",&t);    for(int step=1;step<=t;step++)    {        flag=0;        memset(mmap,0,sizeof(mmap));        for(int i=0;i<=10;i++)            for(int j=0;j<=10;j++)                for(int l=0;l<=30;l++)                    book[i][j][l]=0;        scanf("%d%d%d",&n,&m,&k);        for(int i=0;i<=26;i++)        {            cnt[i]=0;        }        for(int i=1;i<=k;i++)        {            scanf("%d",&cnt[i]);        }        num=n*m;        printf("Case #%d:\n",step);        dfs(1,1);        if(flag==0)        printf("NO\n");    }}