poj 3067 Japan

G - 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

 再提供一组数据：

Sample Input

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

Sample Output

Test case 1: 7

题意：有两排城市，这两排之间有一些城市之间有连接的道路，给出所有道路，问有多少道路是相交的。

分析：求逆序数。我们先把所有的道路按照a升序，a相同时b升序的方法排列。这样从头至尾便利，对于每条道路，我们只需要知道它之前有多少道路的b大于等于它的b就可以了，所以我们只要知道前面有多少b小于它的再用下标减去就可以了。而这个求有多少小于的过程就用树状数组来实现。我们每看到一条边，就把它的b作为下标，把树状数组对应位进行修改。这样一来，树状数组所有下标小于该道路的b的数的总和就是我们要求的b小于该道路的道路数。

解题代码：

 

// File Name: /media/文档/源程序/实验室/poj3067.cpp
// Author: sheng
// Created Time: 2013年03月28日 星期四 13时12分48秒

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

typedef long long LL;
#define Max 2005
int maxy, maxx, point[Max];

struct node
{
    int x;
    int y;
}highway[Max*Max];

bool cmp(const node &a, const node &b)
{
    if (a.x != b.x)
        return a.x < b.x;
    else  return a.y < b.y;
}
int lowbit(int xx)
{
    return xx&(-xx);
}

void updata(int i)
{
    while (i <= maxy)
    {
        point[i] += 1;
        i += lowbit(i);
    }
}

LL add(int i)
{
    LL sum = 0;
    while (i > 0)
    {
        sum += point[i];
        i -= lowbit(i);
    }
    return sum;
}

int main()
{
    int t, k;
    //freopen("t.txt", "r", stdin);
    scanf ("%d",&t);
    for (int i = 0; i < t; i ++)
    {
        memset( point, 0, sizeof (point) );
        scanf ("%d%d%d", &maxx, &maxy, &k);
        for (int j = 0; j < k; j ++)
        {
            scanf("%d%d", &highway[j].x, &highway[j].y);
        }
        sort (highway, highway + k, cmp);
        LL ans = 0;
        for (int j = 0; j < k; j ++ )
        {
            ans += j - add(highway[j].y);//此add目的是为了计算不大于当前 Y 的y值个数然后j - 这个数就是比 Y 大的，也就是线段交点的数
            updata(highway[j].y);
        }
        printf("Test case %d: %lld\n", i + 1, ans);
    }
}