POJ3067 Japan

Posted 2020-08-06 SilverNebula

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 26270		Accepted: 7132

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

Source

Southeastern Europe 2006

 

设每条公路连接左边城市x和右边城市y，按第一关键字x升序，第二关键字y升序排列后，求逆序对即可。

 

 1 /**/
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cmath>
 5 #include<cstring>
 6 #include<algorithm>
 7 using namespace std;
 8 const int mxn=20000;
 9 struct edge{
10     int x,y;
11 }e[mxn*100];
12 int cmp(edge a,edge b){
13     if(a.x!=b.x)return a.x<b.x;
14     return a.y<=b.y;
15 }
16 long long t[mxn];
17 int n,m,k;
18 int a[mxn];
19 inline int lowbit(int x){
20     return x&-x;
21 }
22 void add(int p,int v){
23     while(p<=m){
24         t[p]+=v;
25         p+=lowbit(p);
26     }
27     return;
28 }
29 int sum(int p){
30     int res=0;
31     while(p){
32         res+=t[p];
33         p-=lowbit(p);
34     }
35     return res;
36 }
37 int main(){
38     int T;
39     scanf("%d",&T);
40     int i,j;
41     int cas=0;
42     while(T--){
43         long long ans=0;
44         memset(t,0,sizeof t);
45         scanf("%d%d%d",&n,&m,&k);
46         for(i=1;i<=k;i++) scanf("%d%d",&e[i].x,&e[i].y);
47         sort(e+1,e+k+1,cmp);
48         for(i=1;i<=k;i++){
49             ans+=sum(m)-sum(e[i].y);
50             add(e[i].y,1);
51         }
52         printf("Test case %d: %lld\n",++cas,ans);
53     }
54     return 0;
55 }