POJ 3067 Japan

最新推荐文章于 2019-08-21 16:15:37 发布

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最新推荐文章于 2019-08-21 16:15:37 发布

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原文链接：http://www.cnblogs.com/lzmfywz/archive/2013/06/01/3112957.html

本文解决了一个关于计算两条高速公路交叉次数的问题。通过使用树状数组来求解逆序数的方法，文章详细介绍了输入输出格式、样例及核心算法实现。特别地，文章提供了完整的C++代码实现。

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Japan

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16884		Accepted: 4530

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
题目大意：树状数组求逆序数

#include <stdio.h>
#include <cstring>
#include <algorithm>
using namespace std;
int M, N, K;
long long t[1010];

typedef struct node
{
    int x;
    int y;
}City;

City citys[1000010];

bool cmp(const City &city1, const City &city2)
{
    if (city1.x == city2.x)
    {
        return city1.y < city2.y;
    }
    return city1.x < city2.x;
}

int lowbit(int x)
{
    return x & -x;
}

void add(int pos, long long x)
{
    while(pos <= N)
    {
        t[pos] += x;
        pos += lowbit(pos);
    }
}

int getsum(int pos)
{
    int sum = 0;
    while(pos > 0)
    {
        sum += t[pos];
        pos -= lowbit(pos);
    }
    return sum;
}

int main()
{
    long long result = 0;
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
    {
        result = 0;
        scanf("%d%d%d", &M, &N, &K);
        for (int j = 0; j < K; j++)
        {
            scanf("%d%d", &citys[j].x, &citys[j].y);
        }
        sort(citys, citys + K, cmp);
        memset(t, 0, sizeof(t));
        for (int j = 0; j < K; j++)
        {
            result += getsum(N) - getsum(citys[j].y);
            add(citys[j].y, 1);
        }
        printf("Test case %d: %I64d\n", i, result);  
    }
    return 0;
}