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Black And White
Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
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：有一个n*m个地图，用k中颜色来进行填充，每种颜色可以使用的次数为ci次，∑ci=n*m,要求相邻的格子的颜色不能相同，问是否存在满足要求的染色方案，若存在，则输出其中一种。
分析：注意到n,m≤5,图较小，考虑用dfs来搞，但是光是dfs会T，所以需要加上一个剪枝。
若当前某种颜色的剩余数目大于剩余格子数目的一半，则必定不能完成填充方案，直接跳出。
1 //gaoshenbaoyou ------ pass system test
2 #include <iostream>
3 #include <sstream>
4 #include <ios>
5 #include <iomanip>
6 #include <functional>
7 #include <algorithm>
8 #include <vector>
9 #include <string>
10 #include <list>
11 #include <queue>
12 #include <deque>
13 #include <stack>
14 #include <set>
15 #include <map>
16 #include <cstdio>
17 #include <cstdlib>
18 #include <cmath>
19 #include <cstring>
20 #include <climits>
21 #include <cctype>
22 using namespace std;
23 #define XINF INT_MAX
24 #define INF 0x3FFFFFFF
25 #define MP(X,Y) make_pair(X,Y)
26 #define PB(X) push_back(X)
27 #define REP(X,N) for(int X=0;X<N;X++)
28 #define REP2(X,L,R) for(int X=L;X<=R;X++)
29 #define DEP(X,R,L) for(int X=R;X>=L;X--)
30 #define CLR(A,X) memset(A,X,sizeof(A))
31 #define IT iterator
32 typedef long long ll;
33 typedef pair<int,int> PII;
34 typedef vector<PII> VII;
35 typedef vector<int> VI;
36 const int maxn=30;
37 int a[maxn];
38 bool flag=0;
39 int ans[10][10];
40 int n,m,k;
41 void dfs(int x,int y,int left)
42 {
43 if(!left)
44 {
45 flag=1;
46 return;
47 }
48 for(int i=1;i<=k;i++)
49 if(a[i]>(left+1)/2)return;
50 for(int i=1;i<=k;i++)
51 {
52 if(!a[i])continue;
53 if(x&&ans[x-1][y]==i)continue;
54 if(y&&ans[x][y-1]==i)continue;
55 a[i]--;
56 ans[x][y]=i;
57 if(y<m-1)dfs(x,y+1,left-1);
58 else dfs(x+1,0,left-1);
59 if(flag)return;
60 a[i]++;
61 }
62 return;
63 }
64 int main()
65 {
66 //ios::sync_with_stdio(false);
67 int t;
68 scanf("%d",&t);
69 int cas=1;
70 while(t--)
71 {
72 flag=0;
73 scanf("%d%d%d",&n,&m,&k);
74 int sum=0;
75 int maxx=0;
76 int tot=n*m;
77 for(int i=1;i<=k;i++)
78 scanf("%d",&a[i]);
79 printf("Case #%d:\n",cas++);
80 dfs(0,0,tot);
81 if(flag)
82 {
83 printf("YES\n");
84 for(int i=0;i<n;i++)
85 {
86 for(int j=0;j<m;j++)
87 {
88 if(j)printf(" ");
89 printf("%d",ans[i][j]);
90 }
91 printf("\n");
92 }
93 }
94 else
95 printf("NO\n");
96 }
97 return 0;
98 }
代码君