poj 3067 japan

B - Japan

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit  Status

Description

Japan plans to welcome the ACM ICPC World Finals and a lot of roads must be built for the venue. Japan is tall island with N cities on the East coast and M cities on the West coast (M <= 1000, N <= 1000). K superhighways will be build. Cities on each coast are numbered 1, 2, ... from North to South. Each superhighway is straight line and connects city on the East coast with city of the West coast. The funding for the construction is guaranteed by ACM. A major portion of the sum is determined by the number of crossings between superhighways. At most two superhighways cross at one location. Write a program that calculates the number of the crossings between superhighways.

Input

The input file starts with T - the number of test cases. Each test case starts with three numbers – N, M, K. Each of the next K lines contains two numbers – the numbers of cities connected by the superhighway. The first one is the number of the city on the East coast and second one is the number of the city of the West coast.

Output

For each test case write one line on the standard output: 
Test case (case number): (number of crossings)

Sample Input

1
3 4 4
1 4
2 3
3 2
3 1

Sample Output

Test case 1: 5
思路：
按第一维升序，第二维升序排列之后，用第二维进行树状数组就可以了
当前为 x,y ，要有交点的话,要找出在这个点之前有多少个点的y值大于y,也就是这条边与之前的那些边的交点个数，树状数组向下更新，向上求和

 1 #include<iostream>
 2 #include<string>
 3 #include<cstdio>
 4 #include<vector>
 5 #include<queue>
 6 #include<stack>
 7 #include<algorithm>
 8 #include<cstring>
 9 #include<stdlib.h>
10 #include<string>
11 #include<cmath>
12 using namespace std;
13 #define pb push_back
14 int n,m,k;
15 __int64 p[1100];
16 struct node{
17     int x,y;
18 }num[1001000];
19 int cmp(node a,node b){
20     if(a.x==b.x) return a.y<b.y;
21     return a.x<b.x;
22 }
23 void update(int pos){
24     while(pos>=1){
25         p[pos]+=1;
26         pos-=pos&(-pos);
27     }
28 }
29 __int64 getnum(int pos){
30     __int64 sum=0;
31     while(pos<=m){
32         sum+=p[pos];
33         pos+=pos&(-pos);
34     }
35     return sum;
36 }
37 int main(){
38     int t,cas=0;cin>>t;
39     while(t--){
40         cin>>n>>m>>k;
41         memset(p,0,sizeof(p));
42         for(int i=1;i<=k;i++)
43             scanf("%d%d",&num[i].x,&num[i].y);
44         sort(num+1,num+1+k,cmp);
45         __int64 ans=0;
46         for(int i=1;i<=k;i++){
47             ans+=getnum(num[i].y+1);
48             update(num[i].y);
49         }
50         printf("Test case %d: %I64d\n",++cas,ans);
51     }
52 }

 

转载于:https://www.cnblogs.com/ainixu1314/p/3888693.html