HDU 5113 Black And White(暴力dfs+減枝)

畫船聽雨發表於2015-01-17

題目大意：給你一個n×m的矩陣，然後給你k種顏色，每種顏色有x種，所有的個數加起來恰好為n×m個。問你讓你對這個矩陣進行染色問你，能不能把所有的小方格都染色，而且相鄰兩個顏色不同。

思路：一開始想的是構造，先按照個數進行排序，列舉每一個位置，貪心的策略先放多的，如果可以全部放下就輸出YES，以及存貯的方案，否則輸出NO，但是有bug，一直不對。。。

正解：dfs暴力列舉每一個點，裸的話需要25！，顯然會超時，需要先排個序用構造的策略，讓多的先放這樣可以減枝。然後再dfs就可以了。

Black And White

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 800    Accepted Submission(s): 203
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




 


Source


2014ACM/ICPC亞洲區北京站-重現賽（感謝北師和上交）



#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-9
///#define M 1000100
///#define LL __int64
#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)
#define mod 1000000007
 
using namespace std;
 
const int maxn = 55;
 
 
int mp[10][10];
int n, m, k;
 
 
struct node
{
    int pos;
    int num;
} f[maxn];
 
 
inline int read()
{
    char ch;
    bool flag = false;
    int a = 0;
    while(!((((ch = getchar()) >= '0') && (ch <= '9')) || (ch == '-')));
    if(ch != '-')
    {
        a *= 10;
        a += ch - '0';
    }
    else
    {
        flag = true;
    }
    while(((ch = getchar()) >= '0') && (ch <= '9'))
    {
        a *= 10;
        a += ch - '0';
    }
    if(flag)
    {
        a = -a;
    }
    return a;
}
void write(int a)
{
    if(a < 0)
    {
        putchar('-');
        a = -a;
    }
    if(a >= 10)
    {
        write(a / 10);
    }
    putchar(a % 10 + '0');
}
 
bool dfs(int x, int y)
{
    if(x == n) return true;
    if(y == m) return dfs(x+1, 0);
    for(int i = 0; i < k; i++)
    {
        if(!f[i].num) continue;
        if(x && mp[x-1][y] == i) continue;
        if(y && mp[x][y-1] == i) continue;
        mp[x][y] = i;
        f[i].num--;
        if(dfs(x, y+1)) return true;
        f[i].num++;
    }
    return false;
}
 
bool cmp(node a, node b)
{
    return a.num > b.num;
}
 
int main()
{
    int T;
    scanf("%d", &T);
    int Case = 1;
    while(T--)
    {
        n = read();
        m = read();
        k = read();
        int flag = 0;
        int xp = (n*m+1)/2;
        for(int i = 0; i < k; i++)
        {
            f[i].num = read();
            f[i].pos = i+1;
            if(f[i].num > xp) flag = 1;
        }
        printf("Case #%d:\n",Case++);
        sort(f, f+k, cmp);
        if(flag)
        {
            puts("NO");
            continue;
        }
        dfs(0, 0);
        puts("YES");
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < m-1; j++) printf("%d ",f[mp[i][j]].pos);
            printf("%d\n",f[mp[i][m-1]].pos);
        }
        continue;
    }
}
 








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