Japan - POJ 3067 排序+树状数组

最新推荐文章于 2019-08-09 08:51:54 发布

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本文介绍了一道ACM竞赛题目，任务是计算一组连接城市东海岸与西海岸的高速公路之间的交叉点数量。通过使用树状数组进行排序的方法解决了这个问题，并提供了完整的C++代码实现。

Japan

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19689		Accepted: 5331

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

Sample Output

Test case 1: 5

题意：每次给你n组数，问你这些连线有多少交叉的点。

思路：排序后用树状数组。二元素排序的时候让第一元素从小到大排，第二元素从大到小排，排完后自己画一下就明白了。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int a[10100];
struct node
{ int x,y;
} city[1000005];
bool cmp(node a,node b)
{ if(a.x>b.x)
   return true;
  else if(a.x==b.x)
   return a.y>b.y;
  else
   return false;
}
int lowbit(int x)
{ return x&(-x);
}
void update(int x)
{ while(x<=1000)
  { a[x]++;
    x+=lowbit(x);
  }
}
int sum(int x)
{ int ret=0;
  while(x)
  { ret+=a[x];
    x-=lowbit(x);
  }
  return ret;
}
int main()
{ int t,T,n,m,k,i,j;
  long long ret;
  scanf("%d",&T);
  for(t=1;t<=T;t++)
  { scanf("%d%d%d",&n,&m,&k);
    for(j=1;j<=k;j++)
     scanf("%d%d",&city[j].x,&city[j].y);
    ret=0;
    sort(city+1,city+1+k,cmp);
    memset(a,0,sizeof(a));
    for(j=1;j<=k;j++)
    { ret+=sum(city[j].y-1);
      update(city[j].y);
    }
    printf("Test case %d: %lld\n",t,ret);
  }
}