hdu 4349

http://acm.hdu.edu.cn/showproblem.php?pid=5310 水。 不要用cin. 代码实现 1 2 /************************************************************************* 3 > File Name: code/bc/#ann/1001.cpp 4 > Author: 111qqz 5 > Email: rkz2013@126.com 6 > Created Time: 2015年07月25日 星期六 18时54分24秒 7 ************************************************************************/ 8 9 #include<iostream> 10 #include<iomanip> 11 #include<cstdio> 12 #include<algorithm> 13 #include<cmath> 14 #include<cstring> 15 #include<string> 16 #include<map> 17 #include<set> 18 #include<queue> 19 #include<vector> 20 #include<stack> 21 #define y0 abc111qqz 22 #define y1 hust111qqz 23 #define yn hez111qqz 24 #define j1 cute111qqz 25 #define tm crazy111qqz 26 #define lr dying111qqz 27 using namespace std; 28 #define REP(i, n) for (int i=0;i<int(n);++i) 29 typedef long long LL; 30 typedef unsigned long long ULL; 31 int n,m,p,q; 32 int main() 33 { 34 int T; 35 cin>>T; 36 int ans = 0; 37 while (T--) 38 { 39 // scanf("%d %d %d %d",&n,&m,&p,&q); 40 scanf("%d %d %d %d",&n,&m,&p,&q); 41 ans = n*p; 42 ans = min(ans,n/m*q+n%m*p); 43 ans = min(ans,((n-1)/m+1)*q); 44 printf("%d\n",ans); 45 } 46 47 return 0; 48 }

wa了两次，原因是在同一个点可能有多个基地。。。 所以用set 是错误的，应该用multiset 然后因为这道题看到了map+set实现离散化的另外一种写法 我的代码： 代码实现 1/************************************************************************* 2 > File Name: code/hdoj/4022.cpp 3 > Author: 111qqz 4 > Email: rkz2013@126.com 5 > Created Time: 2015年08月01日 星期六 04时37分20秒 6 ************************************************************************/ 7 8#include<iostream> 9#include<iomanip> 10#include<cstdio> 11#include<algorithm> 12#include<cmath> 13#include<cstring> 14#include<string> 15#include<map> 16#include<set> 17#include<queue> 18#include<vector> 19#include<stack> 20#define y0 abc111qqz 21#define y1 hust111qqz 22#define yn hez111qqz 23#define j1 cute111qqz 24#define tm crazy111qqz 25#define lr dying111qqz 26using namespace std; 27#define REP(i, n) for (int i=0;i<int(n);++i) 28typedef long long LL; 29typedef unsigned long long ULL; 30const int N=2E5+7; 31const int inf = 0x7fffffff; 32 33map<int,int>xmap,ymap; 34multiset<int>x[N]; 35multiset<int>y[N]; 36int main() 37{ 38 int n,m; 39 while (scanf("%d %d",&n,&m)!=EOF) 40 { 41 if (n==0&&m==0) break; 42 for ( int i = 1 ; i <= n ; i++) 43 { 44 x[i].clear(); 45 y[i].clear(); 46 } 47 xmap.clear(); 48 ymap.clear(); 49 int tx,ty; 50 int cntx=0,cnty=0; 51 for ( int i = 1 ; i <= n ; i++ ) 52 { 53 scanf("%d %d",&tx,&ty); 54 if (!xmap[tx]) xmap[tx]=++cntx; 55 if (!ymap[ty]) ymap[ty]=++cnty; 56 x[xmap[tx]].insert(ymap[ty]); 57 y[ymap[ty]].insert(xmap[tx]); 58 } 59 int c,d; 60 set<int>::iterator it; 61 for ( int i = 1; i <= m ; i++ ) 62 { 63 scanf("%d %d",&c,&d); 64 if (c==0) 65 { 66 cout<<x[xmap[d]].size()<<endl; 67 for ( it = x[xmap[d]].begin();it!=x[xmap[d]].end();it++) 68 { 69 y[*it].erase(xmap[d]); 70 } 71 x[xmap[d]].clear(); 72 } 73 else 74 { 75 cout<<y[ymap[d]].size()<<endl; 76 for ( it =y[ymap[d]].begin();it!=y[ymap[d]].end();it++) 77 { 78 x[*it].erase(ymap[d]); 79 80 } 81 y[ymap[d]].clear(); 82 } 83 } 84 printf("\n"); 85 } 86 87 return 0; 88}

Collision Detection # **Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1207 Accepted Submission(s): 367 ** Problem Description In physical simulations, video games and computational geometry, collision detection involves algorithms for checking for collision, i.e. intersection, of two given objects. Collision detection algorithms are a basic component of 3D video games. Without them, characters could go through walls and other obstacles. Here comes an interesting problem, given a ball and a cuboid, you need to detect whether they collide. We say that two objects collide if and only if they share at least one point.