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)
N
Input
TOutput
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,,
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <algorithm> 3 #include <cstdio> 4 #include <cmath> 5 #include <cstring> 6 #include <queue> 7 #include <stack> 8 #include <map> 9 #include <vector> 10 #include <cstdlib> 11 #include <string> 12 13 #define PI acos((double)-1) 14 #define E exp(double(1)) 15 using namespace std; 16 17 vector<pair<long long,long long > >p; 18 long long a[2000000+5]; 19 long long temp[2000000+5]; 20 long long cnt=0;//逆序对的个数 21 void merge(int left,int mid,int right) 22 { 23 int i=left,j=mid+1,k=0; 24 while (( i<=mid )&& (j<=right)) 25 if (a[i]<=a[j]) temp[k++]=a[i++]; 26 else 27 { 28 cnt+=mid+1-i;//关键步骤 29 temp[k++]=a[j++]; 30 } 31 while (i<=mid) temp[k++]=a[i++]; 32 while (j<=right) temp[k++]=a[j++]; 33 for (i=0,k=left; k<=right;) a[k++]=temp[i++]; 34 } 35 void mergeSort(int left,int right) 36 { 37 if (left<right) 38 { 39 int mid=(left+right)/2; 40 mergeSort(left, mid); 41 mergeSort(mid+1, right); 42 merge(left, mid, right); 43 } 44 } 45 int main (void) 46 { 47 int n,u,v,t; 48 cin>>t; 49 for(int i=1;i<=t;i++) 50 { 51 p.clear(); 52 cnt=0; 53 cin>>u>>v>>n; 54 for(int i=0; i<n; i++) 55 { 56 long long k,b; 57 scanf("%lld%lld",&k,&b); 58 p.push_back(make_pair(k,b)); 59 } 60 sort(p.begin(),p.end()); 61 for(int i=0; i<n; i++) 62 a[i]=p[i].second; 63 mergeSort(0,n-1); 64 printf("Test case %d: %lld\n",i,cnt); 65 } 66 return 0; 67 }
作者:weeping
出处:www.cnblogs.com/weeping/
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