poj 3067 Japan

Japan

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14697		Accepted: 3945

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

（与stars极其相似，唯一的不同是上题诗统计之前的个数，而这个统计之后的个数，当然我们可以用当前总数i减去之前的数即可得到。此题bt之处在于不能用long long 只能用__int64。）

题意：顺序给两组平行的点依次编号1~N和1~M，给定K个线段在两组点之间，求相交(cross)的线段对有多少个，同一个起点或终点不算相交。

思路：树状数组，首先对x进行排序，x相等则按y进行排序，排序后以y作为树状数组。当有一条连线时，我们知道与这条连线有交点的情况有两种，(a[i].x>a[j].x&&a[i].y<a[j].y) 和 (a[i].x<a[j].x&&a[i].y>a[j].y) ,但是我们已经将数据排好序了，那么第二种情况就可以不用在当前考虑，我们只要知道a[i].y 到 m之间有多少条连线即可，与第二种情况相连的情况由a[j].y去计算。
于是计算当前这条连线有多少个交叉点即：getsum(m)-getsum(a[i].y)。

 

AC代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define maxn 2000
using namespace std;
#define ll __int64
ll c[maxn*maxn];
struct node
{
 int x,y;
}nn[maxn*maxn];
int n,m,k;
bool cmp( node a,node b)
{
 return a.x==b.x?a.y<b.y:a.x<b.x;
}
int lowbit( int i)
{
 return i&(-i);
}
void updata( int i)
{
  while(i<=m)
  {
 c[i]+=1;
 i+=lowbit(i);
  }
}
ll sum( int i)
{
 ll s=0;
 while(i>0)
 {
 s+=c[i];
 i-=lowbit(i);
 }
 return s;
}
int main( )
{
  int test;
  int cases=0;
  scanf("%d",&test);
  while(test--)
  {
 memset(c,0,sizeof(c));
 scanf("%d%d%d",&n,&m,&k);
 for( int i=1;i<=k;i++)
 {
   scanf("%d%d",&nn[i].x,&nn[i].y);
 }
 sort(nn+1,nn+k+1,cmp);
 ll ans=0;
 for( int i=1;i<=k;i++)
 {
    ans+=sum(m)-sum(nn[i].y);
    updata(nn[i].y);
 }
 printf("Test case %d: %I64d\n",++cases,ans);
  }
  return 0;
}

链接：http://poj.org/problem?id=3067