cdojR - Japan

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[+];
 long long  temp[+];
 long long  cnt=;//逆序对的个数
 void merge(int left,int mid,int right)
 {
     int i=left,j=mid+,k=;
     while (( i<=mid )&& (j<=right))
         if (a[i]<=a[j]) temp[k++]=a[i++];
         else
         {
             cnt+=mid+-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=,k=left; k<=right;) a[k++]=temp[i++];
 }
 void mergeSort(int left,int right)
 {
     if (left<right)
     {
         int mid=(left+right)/;
         mergeSort(left, mid);
         mergeSort(mid+, right);
         merge(left, mid, right);
     }
 }
 int main (void)
 {
     int n,u,v,t;
     cin>>t;
     for(int i=;i<=t;i++)
     {
         p.clear();
         cnt=;
         cin>>u>>v>>n;
         for(int i=; 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=; i<n; i++)
             a[i]=p[i].second;
         mergeSort(,n-);
         printf("Test case %d: %lld\n",i,cnt);
     }
     return ;
 }