POJ 3067 Japan (树状数组)

Japan

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16682		Accepted: 4467

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

1
3 4 4
1 4
2 3
3 2
3 1

Sample Output

Test case 1: 5

Source

Southeastern Europe 2006

题意：东西两岸有N，M座城市 从北到南标号分别为1...N or M 在它们之间建铁路，每个站最多只能有两条铁路 求总的交点数

思路：各连线按x降序（x相同，y降序）重排 再以y 建树状数组 注意有重边的情况

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

const int N=1010;

struct node{
    int x,y;
}tree[N*N];

long long array[N];

int cmp(node a,node b){
    if(a.x!=b.x)
        return a.x>b.x;
    return a.y>b.y;
}

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

void add(int rt){
    while(rt<N){
        array[rt]+=1;
        rt+=lowbit(rt);
    }
}

long long sum(int rt){
    long long ans=0;
    while(rt>0){
        ans+=array[rt];
        rt-=lowbit(rt);
    }
    return ans;
}

int main(){

    //freopen("input.txt","r",stdin);

    int t,n,m,k;
    int cases=0;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%d",&n,&m,&k);
        memset(array,0,sizeof(array));
        for(int i=0;i<k;i++)
            scanf("%d%d",&tree[i].x,&tree[i].y);
        sort(tree,tree+k,cmp);
        long long ans=0;
        for(int i=0;i<k;i++){
            ans+=sum(tree[i].y-1);
            add(tree[i].y);
        }
        printf("Test case %d: %I64d\n",++cases,ans);
    }
    return 0;
}