BNUOJ 33898 Cannon
Cannon
This problem will be judged on HDU. Original ID: 4499
64-bit integer IO format: %I64d Java class name: Main
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
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
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
Source
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
int n,m,k,ans;
int mp[][];
void dfs(int cur,int has){
int x = cur/m;
int y = cur%m;
if(has > ans) ans = has;
if(cur == n*m) return;
if(mp[x][y]) {dfs(cur+,has);return;}
dfs(cur+,has);
int i,cnt = ;
for(i = x-; i >= ; i--){
if(mp[i][y]) cnt++;
if(cnt == ) break;
}
if(cnt == && mp[i][y] == ) return;
cnt = ;
for(i = y-; i >= ; i--){
if(mp[x][i]) cnt++;
if(cnt == ) break;
}
if(cnt == && mp[x][i] == ) return;
mp[x][y] = ;
dfs(cur+,has+);
mp[x][y] = ;
}
int main() {
int i,j,x,y;
while(~scanf("%d %d %d",&n,&m,&k)){
memset(mp,,sizeof(mp));
for(i = ; i < k; i++){
scanf("%d %d",&x,&y);
mp[x][y] = ;
}
ans = ;
dfs(,);
printf("%d\n",ans);
}
return ;
}
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