HDU 5113 Black And White （搜索DFS）2014ICPC北京站现场赛

Black And White

                                                                       Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
                                                                                                     Total Submission(s): 1869    Accepted Submission(s): 512
                                                                                                                                           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

 

题意：在一个N*M大小的方格中让你随机的涂满K种颜色，第i种颜色需要涂Ci个格子，满足所有颜色一定能涂满整个方格。

分析：在做这道题的时候有两种想法，一种是构造，一种是搜索，构造有点复杂，又考虑到N最大为5，然后果断选择了搜索。但是搜索的时候一定要进行剪枝，否则一定会TLE。当然剪枝也是有技巧的。

#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <string>#include <vector>#include <deque>#include <list>#include <set>#include <map>#include <stack>#include <queue>#include <cctype>#include <numeric>#include <iomanip>#include <bitset>#include <sstream>#include <fstream>#define debug "output for debug\n"#define pi (acos(-1.0))#define eps (1e-8)#define inf 0x3f3f3f3f#define ll long long int#define lson l , m , rt << 1#define rson m + 1 , r , rt << 1 | 1using namespace std;const int mod = 1000000007;const int Max = 100005;int t,n,m,k;int col[30];int mp[30][30];int flag;bool check(int x,int y,int color){    if(x-1>0&&mp[x-1][y]==color)        return false;    if(y-1>0&&mp[x][y-1]==color)        return false;    return true;}bool dfs(int x,int y){    int tmp=(n*m-((x-1)*m+y-1)+1)/2;///在这里进行了剪枝，比在选颜色的for循环里剪要优化很多，自己可以试试，另外如果对颜色数组按照大小排序会更快。    for(int i=1;i<=k;i++)    {        if(tmp<col[i])            return false;    }    for(int i=1; i<=k; i++)    {        if(col[i]>0&&check(x,y,i))        {            col[i]--;            mp[x][y]=i;            if(x==n&&y==m)                return true;            if(y+1<=m)            {                if(dfs(x,y+1))                    return true;            }            else            {                if(x+1<=n)                {                    if(dfs(x+1,1))                        return true;                }            }            col[i]++;            mp[x][y]=0;        }    }    return false;}int main(){    cin>>t;    int cnt=1;    while(t--)    {        scanf("%d%d%d",&n,&m,&k);        memset(mp,0,sizeof mp);        flag=0;        int ma=0;        for(int i=1; i<=k; i++)        {            scanf("%d",&col[i]);            ma=max(ma,col[i]);        }        printf("Case #%d:\n",cnt++);        if(ma>(n*m+1)/2)        {            puts("NO");            continue;        }        if(dfs(1,1))        {            puts("YES");            for(int i=1; i<=n; i++)            {                for(int j=1; j<m; j++)                    printf("%d ",mp[i][j]);                printf("%d\n",mp[i][m]);            }        }        else            puts("NO\n");    }    return 0;}

题目链接：点击打开链接