Cannon

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. 

SampleInput

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 

SampleOutput

8 
9

题意就是给你一个n*m的棋盘，然后上面已经有了 棋子，并给出这些棋子的坐标，但是这些棋子是死的就是不能动，然后让你在棋盘上面摆炮，但是炮之间不能互相吃，吃的规则我们斗懂得 炮隔山打嘛，问你最多能放几个炮

法一：并查集

#include <iostream>
#include <stack>
#include <stdio.h>
using namespace std;
const int MAX_N = 10000 + 100;
const int MAX_M = 100000 + 100;
int p[MAX_N];
int _find(int x)
{
    return p[x] == x ? x : (p[x] = _find(p[x]));
}
int n, m;
stack <int> s;
struct Edge
{
    int u, v;
};
Edge edge[MAX_M];
int res[MAX_M];

int main()
{
    while(scanf("%d%d", &n, &m) != EOF)
    {
        for(int i = 0; i < n; i++)
            p[i] = i;
        for(int i = 0; i < m; i++)
            scanf("%d%d", &edge[i].u, &edge[i].v);
        res[m - 1] = n;
        for(int i = m - 1; i > 0; i--)
        {
            int a = _find(edge[i].u);
            int b = _find(edge[i].v);
            if(a != b)
            {
                p[a] = b;
                res[i - 1] = res[i] - 1;
            }
            else
                res[i - 1] = res[i];
        }
        for(int i = 0; i < m; i++)
            printf("%d ", res[i]);
    }
    return 0;
}

法二，DFS

数据范围很小，明显是搜索。

主要剪枝，就是不要和前面的冲突了、

/* ***********************************************
Author        :kuangbin
Created Time  :2013/8/24 14:38:00
File Name     :F:2013ACM练习比赛练习2013通化邀请赛1007.cpp
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
int n,m;
int g[10][10];
int ans ;

void dfs(int x,int y,int cnt)
{
    if(x >= n)
    {
        ans = max(ans,cnt);
        return;
    }
    if(y >= m)
    {
        dfs(x+1,0,cnt);
        return;
    }
    if(g[x][y] == 1)
    {
        dfs(x,y+1,cnt);
        return;
    }
    dfs(x,y+1,cnt);
    bool flag = true;
    int t;
    for(t = x-1;t >= 0;t--)
        if(g[t][y])
        {
            break;
        }
    for(int i = t-1;i >= 0;i--)
        if(g[i][y])
        {
            if(g[i][y]==2)flag = false;
            break;
        }
    if(!flag)return;
    for(t = y-1;t >= 0;t--)
        if(g[x][t])
            break;
    for(int j = t-1;j >= 0;j--)
        if(g[x][j])
        {
            if(g[x][j] == 2)flag = false;
            break;
        }
    if(!flag)return;
    g[x][y] = 2;
    dfs(x,y+1,cnt+1);
    g[x][y] = 0;
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int Q;
    int u,v;
    while(scanf("%d%d%d",&n,&m,&Q) == 3)
    {
        memset(g,0,sizeof(g));
        while(Q--)
        {
            scanf("%d%d",&u,&v);
            g[u][v] = 1;
        }
        ans = 0;
        dfs(0,0,0);
        printf("%d ",ans);
    }
    return 0;
}