题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4499
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CannonTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 513 Accepted Submission(s): 297
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
Sample Output
Source
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题意:给定一个棋盘，以及棋子的位置，问在该棋盘上最多还能放多少个棋子"炮"，并且要保证"炮"之间不能相互攻击。
题解:可以用搜索做，从左至右，从上至下摆棋子，每次都要判断该点是否可以摆放。
——开始一直WA/TLE 郁闷——又稀里糊涂的过了~~ 果断时间再来看看。
AC代码:
#include<iostream>
#include<cstring>
#define N 6
using namespace std;
int init[N][N];
int n,m,q,tx,ty,res;
bool check(int x,int y){
if(init[x][y])return false;
init[x][y]=2;
for(int i=0;i<y-1;i++){
if(init[x][i]==2){
int k=i+1,flag=0;
while(k<=y){
if(init[x][k]==2&&flag){
init[x][y]=0;return false;
}
if(init[x][k]>0)flag++;
if(flag>1)break;
k++;
}
}
}
for(int i=0;i<x-1;i++){
if(init[i][y]==2){
int k=i+1,flag=0;
while(k<=x){
if(init[k][y]==2&&flag){
init[x][y]=0;return false;
}
if(init[k][y]>0)flag++;
if(flag>1)break;
k++;
}
}
}
init[x][y]=0;
return true;
}
void dfs(int x,int y,int cnt){
if(cnt>res)res=cnt;
if(y==m){
x++;y=0;
if(x>=n)return;
}
for(int i=x;i<n;i++)
for(int j=y;j<m;j++){
if(check(i,j)){
init[i][j]=2;
dfs(i,j+1,cnt+1);
init[i][j]=0;
}
return dfs(i,j+1,cnt);
}
}
int main()
{
cin.sync_with_stdio(false);
while(cin>>n>>m>>q)
{
res=0;
memset(init,0,sizeof(init));
while(q--){
cin>>tx>>ty; init[tx][ty]=1;
}
dfs(0,0,0);
cout<<res<<endl;
}
return 0;
}
原文：http://blog.csdn.net/mummyding/article/details/43526463