Japan
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 20991 | Accepted: 5685 |
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)
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
————————————————分割线————————————————
题目大意:
求k条线段的交点个数
思路:
将西边的城镇从小到大排序,如果编号一样,则对东边的城镇按从小到大排序,求逆序数 (或者按从大到小排序,求比它小的数)
画画图就非常清楚了
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define ll long long
#define M 1000000+10
using namespace std;
int n,m,k;
struct node
{
int x,y;
}a[M];
int c[1010];
bool cmp(node x,node y)
{
return (x.x<y.x||(x.x==y.x&&x.y<y.y));
}
inline void update(int x,int v)
{
while(x<=m){
c[x]+=v;
x+=x&-x;
}
}
inline ll getsum(int x)
{
ll sum=0;
while(x>0){
sum+=c[x];
x-=x&-x;
}
return sum;
}
int main()
{
int T,cas=1;
scanf("%d",&T);
while(T--){
ll cnt=0;
memset(c,0,sizeof(c));
scanf("%d %d %d",&n,&m,&k);
for(int i=0;i<k;++i){
scanf("%d %d",&a[i].x,&a[i].y);
}
sort(a,a+k,cmp);
for(int i=0;i<k;++i){
update(a[i].y,1);
cnt+=i-getsum(a[i].y)+1;
}
printf("Test case %d: %lld\n",cas++,cnt);
}
return 0;
}