Cannon

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

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

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

HINT

题意

在一个象棋棋盘上放炮，要求两个炮不能互相打到，然后问你最多能放几个炮

题解：

直接dfs就好了，范围很小

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 2000001
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar(‘0‘);puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
//**************************************************************************************

int a[30],b[30],visit[7][7],n,m,q,ans;
void dfs(int x,int y,int cnt)//一行一行地搜索，直到找到最后一行时结束时记录最大值
{
    if(x>=n){
        ans=max(ans,cnt);
        return;
    }
    if(y>=m){
        dfs(x+1,0,cnt);
        return;
    }
    if(visit[x][y]){
        dfs(x,y+1,cnt);
        return;
    }
    dfs(x,y+1,cnt);
    int t,flag=0;
    for(t=y-1;t>=0;t--)
        if(visit[x][t]) break;
    for(int i=t-1;i>=0;i--)
    {
        if(visit[x][i]==2) {flag=1;break;}
        if(visit[x][i]) break;
    }
    if(flag)return;//判断这一列上是否存在炮互吃
    for(t=x-1;t>=0;t--)
        if(visit[t][y]) break;
    for(int i=t-1;i>=0;i--){
        if(visit[i][y]==2) {flag=1;break;}
        if(visit[i][y]) break;
    }
    if(flag) return;//判断这一行上是否存在炮互吃
    visit[x][y]=2;
    dfs(x,y+1,cnt+1);
    visit[x][y]=0;//回溯
}
int main()
{
    while(scanf("%d%d%d",&n,&m,&q)!=EOF){
        memset(visit,0,sizeof(visit));
        for(int i=0;i<q;i++){
            scanf("%d%d",&a[i],&b[i]);
            visit[a[i]][b[i]]=1;
        }
        ans=0;
        dfs(0,0,0);
        printf("%d\n",ans);
    }
    return 0;
}