hdu 4499 Cannon 暴力dfs搜索

Cannon

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 589    Accepted Submission(s): 338

Problem Description

In Chinese Chess, there is one kind of powerful chessmen called Cannon. It can move horizontally or vertically along the chess grid. At each move, it can either simply move to another empty cell in the same line without any other chessman along the route or perform an eat action. The eat action, however, is the main concern in this problem. 
An eat action, for example, Cannon A eating chessman B, requires two conditions: 
1、A and B is in either the same row or the same column in the chess grid. 
2、There is exactly one chessman between A and B. 
Here comes the problem. 
Given an N x M chess grid, with some existing chessmen on it, you need put maximum cannon pieces into the grid, satisfying that any two cannons are not able to eat each other. It is worth nothing that we only account the cannon pieces you put in the grid, and no two pieces shares the same cell.

 

Input

There are multiple test cases. 
In each test case, there are three positive integers N, M and Q (1<= N, M<=5, 0<=Q <= N x M) in the first line, indicating the row number, column number of the grid, and the number of the existing chessmen. 
In the second line, there are Q pairs of integers. Each pair of integers X, Y indicates the row index and the column index of the piece. Row indexes are numbered from 0 to N-1, and column indexes are numbered from 0 to M-1. It guarantees no pieces share the same cell.

 

Output

There is only one line for each test case, containing the maximum number of cannons.

 

Sample Input

   
   
   
   

    
    
    
    4 4 2 
1 1 1 2 
5 5 8 
0 0 1 0 1 1 2 0 2 3 3 1 3 2 4 0 
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    8 
9
   
   
   
   

 

链接：http://acm.hdu.edu.cn/showproblem.php?pid=4499

题意：给你一个n*m的棋盘，有q个点已经放了棋。问最多还能放多少炮。要求炮与炮之间不能互相攻击。也就是两个炮直接不能隔着炮或者隔着棋。

做法：棋盘大小最大25个格子。2^25=3*10^7。如果枚举每个状态还要判断可不可行会超时。 所以我用了dfs。每次那个位置如果没有棋，就摆上炮dfs下去，每次摆判断和正上方一排冲突不冲突，还有左边一排冲不冲突。不冲突才可以摆。   再把炮去掉dfs下去。如果没有棋，就dfs不摆炮就行了。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <malloc.h>
#include <ctype.h>
#include <math.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#include <stack>
#include <queue>
#include <vector>
#include <deque>
#include <set>
#include <map>

int n,m,q;
int mp[5][5];
int ass;
int dfs(int nw,int fa)
{
	if(nw==n*m)
	{ 
		return fa;
	}
	int x=nw/m;
	int y=nw%m;
	int maxx=fa;
	if(mp[x][y]!=1)
	{
		int flag=0;
		if(mp[1][0]==2&&mp[3][0]==2&&nw==20)
			int kk=1;
		for(int i=x-1;i>=0;i--)
		{
			if(flag==1&&mp[i][y]==2)
				flag=2;
			if(flag==1&&mp[i][y]==1)
				break;
			if(flag==0&&(mp[i][y]==1||mp[i][y]==2))//炮
				flag=1; 
		}
		int flag2=0;
		for(int i=y-1;i>=0;i--)
		{
			if(flag2==1&&mp[x][i]==2)
				flag2=2;
			if(flag2==1&&mp[x][i]==1)
				break;
			if(flag2==0&&(mp[x][i]==1||mp[x][i]==2))//炮
				flag2=1;  
		}
		if(flag!=2&&flag2!=2)
		{
			mp[x][y]=2;
			maxx=max(maxx,dfs(nw+1,fa+1));
			mp[x][y]=0; 
			maxx=max(maxx,dfs(nw+1,fa));
		} 
		else
			maxx=max(maxx,dfs(nw+1,fa));
	}
	else
		maxx=max(maxx,dfs(nw+1,fa));
	return maxx;
}

int main()
{
	while(scanf("%d%d%d",&n,&m,&q)!=EOF)
	{
		memset(mp,0,sizeof mp);
		ass=0;
		for(int i=0;i<q;i++)
		{
			int x,y;
			scanf("%d%d",&x,&y);
			mp[x][y]=1; 
		} 
		printf("%d\n",dfs(0,0));
	}
	return 0;
}