[树状数组][逆序数]Japan

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

思路：将k条path按path[i].x从小到大排序(若path[i].x相同，则按path[i].y从小到大排序)，此时构成的path[1~k].y序列的逆序数便是答案；树状数组/归并排序求逆序数；
树状数组：https://www.cnblogs.com/hsd-/p/6139376.html
AC代码：

//树状数组求逆序数O(nlogn+logn)
#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define lowbit(x) x&(-x)
typedef long long ll;
using namespace std;

struct Path{
  int x,y;
}path[1000010];

bool cmp(Path a,Path b){
  if(a.x!=b.x) return a.x<b.x;
  else return a.y<b.y;
}

int n,m,k;
int c[1010];

void add(int x,int val){
  for(int i=x;i<=m;i+=lowbit(i)) c[i]+=val;
}

ll getsum(int x){
  ll ret=0;
  for(int i=x;i>0;i-=lowbit(i)) ret+=c[i];
  return ret;
}

int main()
{
    int t;
    scanf("%d",&t);
    for(int s=1;s<=t;s++){
        scanf("%d%d%d",&n,&m,&k);
        for(int i=1;i<=k;i++) scanf("%d%d",&path[i].x,&path[i].y);
        sort(path+1,path+1+k,cmp);
        memset(c,0,sizeof(c));
        ll ans=0;
        for(int i=1;i<=k;i++){
            add(path[i].y,1);//出现过该数，则标记为1
            ans+=(i-getsum(path[i].y));//前i的数里面大于path[i].y的数的个数等于i-前i个数里面小于等于path[i].y的数的个数(即getsum(path[i].y))
        }
        printf("Test case %d: ",s);
        cout<<ans<<endl;
    }
    return 0;
}

AC代码：

//归并排序求逆序数O(nlogn+nlogn)
#include <iostream>
#include<cstdio>
#include<algorithm>
typedef long long ll;
using namespace std;

struct Path{
  int x,y;
}path[1000010];

bool cmp(Path a,Path b){
  if(a.x!=b.x) return a.x<b.x;
  else return a.y<b.y;
}

int n,m,k;
int b[1000010];
int tmp[1000010];
ll ans=0;

void merge_(int l,int m,int r){
  int cnt=0;
  int i,j;
  for(i=l,j=m+1;i<=m&&j<=r;){
     if(b[i]<=b[j]) {tmp[++cnt]=b[i]; i++;}
     else{
        tmp[++cnt]=b[j]; j++;
        ans+=(m-i+1);
     }
  }
  while(i<=m) {tmp[++cnt]=b[i]; i++;}
  while(j<=r) {tmp[++cnt]=b[j]; j++;}
  for(int i=l,j=1;i<=r&&j<=cnt;i++,j++) b[i]=tmp[j];
}

void merge_sort(int l,int r){
  if(l==r) return;
  int m=(l+r)>>1;
  merge_sort(l,m);
  merge_sort(m+1,r);
  merge_(l,m,r);
}

int main()
{
    int t;
    scanf("%d",&t);
    for(int s=1;s<=t;s++){
        scanf("%d%d%d",&n,&m,&k);
        for(int i=1;i<=k;i++) scanf("%d%d",&path[i].x,&path[i].y);
        sort(path+1,path+1+k,cmp);
        for(int i=1;i<=k;i++) b[i]=path[i].y;
        ans=0;
        merge_sort(1,k);
        printf("Test case %d: ",s);
        cout<<ans<<endl;
    }
    return 0;
}

[树状数组][逆序数]Japan

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