Japan POJ - 3067 树状数组

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本文介绍了一个算法问题，即计算日本为迎接ACM-ICPC世界总决赛而计划建设的直连东西海岸城市的K条超级高速公路之间的交叉数量。文章提供了一段C++代码实现，通过排序和树状数组来高效计算总的交叉次数。

Japan

POJ - 3067

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

废话不多说直接上代码

#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
#include <stack>
using namespace std;
const int MAXN = 1000000;
struct Node {
	int u,v;
} eg[MAXN];
int n,m,k;
int c[MAXN];
int lowbit(int x) {
	return x&(-x);
}
bool cmp(Node a,Node b) {
	if(a.u!=b.u) return a.u<b.u;
	return a.v<b.v;
}
long long int sum(int x) {
	long long ans=0;
	while(x) {
		ans+=c[x];
		x-=lowbit(x);
	}
	return ans;
}
void add(int x) {
	while(x<=m) {
		c[x]++;
		x+=lowbit(x);
	}
}
int main() {
	int t;
	scanf("%d",&t);
	int ca=1;
	while(t--) {
		memset(c, 0, sizeof(c));
		scanf("%d %d %d",&n,&m,&k);
		for(int i=0; i<k; i++) {
			scanf("%d %d",&eg[i].u,&eg[i].v);
		}
		sort(eg,eg+k,cmp);
		long long int ans=0;
		for(int i=0; i<k; i++) {
			ans+=sum(m)-sum(eg[i].v);
			add(eg[i].v);
		}
		printf("Test case %d: ",ca);
		ca++;
		printf("%lld\n",ans);
	}


	return 0;
}