传送门:http://acm.hdu.edu.cn/showproblem.php?pid=3682
题意:
给你一个三维的立方体,每次去掉一列,去掉m列之后问你总共去掉了多少个1*1*1的小立方体。
题解:
用hash来标记每个小立方体,用vector保存,最后去掉重复的,余下的即位答案。
AC代码:
Accepted | 3682 | 203MS | 4512K | 2373 B | G++ | XH_Reventon |
#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <cstdlib> #include <cmath> #include <vector> #include <list> #include <deque> #include <queue> #include <iterator> #include <stack> #include <map> #include <set> #include <algorithm> #include <cctype> using namespace std; #define si1(a) scanf("%d",&a) #define si2(a,b) scanf("%d%d",&a,&b) #define sd1(a) scanf("%lf",&a) #define sd2(a,b) scanf("%lf%lf",&a,&b) #define ss1(s) scanf("%s",s) #define pi1(a) printf("%d\n",a) #define pi2(a,b) printf("%d %d\n",a,b) #define mset(a,b) memset(a,b,sizeof(a)) #define forb(i,a,b) for(int i=a;i<b;i++) #define ford(i,a,b) for(int i=a;i<=b;i++) typedef long long LL; const int N=33; const int INF=0x3f3f3f3f; const double PI=acos(-1.0); const double eps=1e-8; vector<int> num; int main() { // freopen("input.txt","r",stdin); int nCase; int n, m; scanf("%d", &nCase); while(nCase--) { num.clear(); int len=0; scanf("%d%d", &n, &m); getchar(); while(m--) { char a, b; int val1, val2; scanf("%c=%d,%c=%d", &a, &val1, &b, &val2); getchar(); if(a=='X') { if(b=='Y') for(int i=1; i<=n; i++) num.push_back(val1*n*n+val2*n+i); else for(int i=1; i<=n; i++) num.push_back(val1*n*n+i*n+val2); } if(a=='Y') { if(b=='X') for(int i=1; i<=n; i++) num.push_back(val2*n*n+val1*n+i); else for(int i=1; i<=n; i++) num.push_back(i*n*n+val1*n+val2); } if(a=='Z') { if(b=='X') for(int i=1; i<=n; i++) num.push_back(val2*n*n+i*n+val1); else for(int i=1; i<=n; i++) num.push_back(i*n*n+val2*n+val1); } } sort(num.begin(), num.end());//排序 num.erase( unique( num.begin(), num.end() ), num.end());//去重 printf("%d\n", num.size()); } return 0; }