HDU 3682 To Be an Dream Architect （hash）

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

題意：

給你一個三維的立方體，每次去掉一列，去掉m列之後問你總共去掉了多少個1*1*1的小立方體。

題解：

用hash來標記每個小立方體，用vector儲存，最後去掉重複的，餘下的即位答案。

AC代碼：

3682

203MS

4512K

2373 B

G++

#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;}