cdojR

地址：http://acm.uestc.edu.cn/#/contest/show/95

题目：

R - Japan

Time Limit: 3000/1000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others)

Submit Status

N

Input

T

Output

For each test case write one line on the standard output: Test case (case number): (number of crossings)

Sample input and output

Sample Input	Sample Output
1 3 4 4 1 4 2 3 3 2 3 1	Test case 1: 5

Hint

The data used in this problem is unofficial data prepared by pfctgeorge. So any mistake here does not imply mistake in the offcial judge data.

思路：

又是逆序对，具体的不多说了，和前面的某题一样，，，

归并求逆序对数，，

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#include <cstdlib>
#include <string>

#define PI acos((double)-1)
#define E exp(double(1))
using namespace std;

vector<pair<long long,long long > >p;
long long  a[2000000+5];
long long  temp[2000000+5];
long long  cnt=0;//逆序对的个数
void merge(int left,int mid,int right)
{
    int i=left,j=mid+1,k=0;
    while (( i<=mid )&& (j<=right))
        if (a[i]<=a[j]) temp[k++]=a[i++];
        else
        {
            cnt+=mid+1-i;//关键步骤
            temp[k++]=a[j++];
        }
    while (i<=mid) temp[k++]=a[i++];
    while (j<=right) temp[k++]=a[j++];
    for (i=0,k=left; k<=right;) a[k++]=temp[i++];
}
void mergeSort(int left,int right)
{
    if (left<right)
    {
        int mid=(left+right)/2;
        mergeSort(left, mid);
        mergeSort(mid+1, right);
        merge(left, mid, right);
    }
}
int main (void)
{
    int n,u,v,t;
    cin>>t;
    for(int i=1;i<=t;i++)
    {
        p.clear();
        cnt=0;
        cin>>u>>v>>n;
        for(int i=0; i<n; i++)
        {
            long long k,b;
            scanf("%lld%lld",&k,&b);
            p.push_back(make_pair(k,b));
        }
        sort(p.begin(),p.end());
        for(int i=0; i<n; i++)
            a[i]=p[i].second;
        mergeSort(0,n-1);
        printf("Test case %d: %lld
",i,cnt);
    }
    return 0;
}