http://acm.hdu.edu.cn/showproblem.php?pid=3682

To Be an Dream Architect

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

The “dream architect” is the key role in a team of “dream extractors” who enter other’s dreams to steal secrets. A dream architect is responsible for crafting the virtual world that the team and the target will dream into. To avoid the target noticing the world
is artificial, a dream architect must have powerful 3D imagination.

Cobb uses a simple 3D imagination game to test whether a candidate has the potential to be an dream architect. He lets the candidate imagine a cube consisting of n×n×n blocks in a 3D coordinate system as Figure 1. The block at bottom left front corner is marked
(1, 1, 1) and the diagonally opposite block is marked (n, n, n). Then he tells the candidate that the blocks on a certain line are eliminated. The line is always parallel to an axis. After m such block eliminations, the candidate is asked to tell how many
blocks are eliminated. Note that one block can only be eliminated once even if it is on multiple lines.

Here is a sample graph according to the first test case in the sample input:

Input

The first line is the number of test cases.

In each test case, the first line contains two integers n and m( 1 <= n <= 1000, 0 <= m <= 1000).，meaning that the cube is n x n x n and there are m eliminations.

Each of the following m lines represents an elimination in the following format:

axis_1=a, axis_2=b

where axis_i (i=1, 2) is ‘X’ or ‘Y’, or ‘Z’ and axis_1 is not equal to axis_2. a and b are 32-bit signed integers.

Output

For each test case output the number of eliminated blocks.

Sample Input

2
3 2
Y=1,Z=3
X=3,Y=1
10 2
X=3,Y=3
Y=3,Z=3

Sample Output

5
19

Source

2010 Asia Hangzhou Regional Contest

题意：每次对立方体切割一条线上的块，问切割了多少个。

思路：暴力。。。把每个小立方块哈希编号为x*n*n+y*n+z，然后最后排序判重下就好。。。。。

stl大法好。。直接用unique。。。

/**
 * @author neko01
 */
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define clr(a) memset(a,0,sizeof a)
#define clr1(a) memset(a,-1,sizeof a)
#define dbg(a) printf("%d\n",a)
typedef pair<int,int> pp;
const double eps=1e-9;
const double pi=acos(-1.0);
const int INF=0x3f3f3f3f;
const LL inf=(((LL)1)<<61)+5;
vector<int>ans;
int main()
{
    int n,m,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        getchar();
        ans.clear();
        while(m--)
        {
            char c1,c2;
            int a,b;
            scanf("%c=%d,%c=%d",&c1,&a,&c2,&b);
            getchar();
            if(c1=='X')
            {
                if(c2=='Y')
                    for(int i=1;i<=n;i++)
                        ans.pb(a*n*n+b*n+i);
                else
                    for(int i=1;i<=n;i++)
                        ans.pb(a*n*n+i*n+b);
            }
            else if(c1=='Y')
            {
                if(c2=='X')
                    for(int i=1;i<=n;i++)
                        ans.pb(b*n*n+a*n+i);
                else
                    for(int i=1;i<=n;i++)
                        ans.pb(i*n*n+a*n+b);
            }
            else
            {
                if(c2=='X')
                    for(int i=1;i<=n;i++)
                        ans.pb(b*n*n+i*n+a);
                else
                    for(int i=1;i<=n;i++)
                        ans.pb(i*n*n+b*n+a);
            }
        }
        sort(ans.begin(),ans.end());
        int sz=unique(ans.begin(),ans.end())-ans.begin();
        printf("%d\n",sz);
    }
    return 0;
}