POJ 3067 Japan

Japan

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 25489		Accepted: 6907

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

 

 

 

解析：树状数组（单点更新，区间查询）。把道路进行排序（按照w从小到大排序，w相同则按e从小到大排序），这样，对于道路r_i，若e_j>e_i（0≤j<i），则r_i、rj两条道路存在交点。通过树状数组，我们可以知道道路r_i前面有多少条道路的e小于等于e_i，再用下标减去它即可得到r_i前面道路的e大于e_i的条数，总和就是结果。

 

 

 

#include <cstdio>
#include <cstring>
#include <algorithm>
#define lowbit(x) (x)&(-x)
#define ll long long
using namespace std;

struct Road{
    int w, e;
    bool operator < (const Road& b)const
    {
        if(w != b.w) return w<b.w;
        return e<b.e;
    }
}r[1000005];

int c[1005];
int n, m, k;

void add(int x, int val)
{
    for(int i = x; i <= 1000; i += lowbit(i))
        c[i] += val;
}

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

void solve()
{
    memset(c, 0, sizeof(c));
    sort(r, r+k);
    ll res = 0;
    for(int i = 0; i < k; ++i){
        res += i-sum(r[i].e);
        add(r[i].e, 1);
    }
    printf("%I64d\n", res);
}

int main()
{
    int t, cn = 0;
    scanf("%d", &t);
    while(t--){
        scanf("%d%d%d", &n, &m, &k);
        for(int i = 0; i < k; ++i)
            scanf("%d%d", &r[i].w, &r[i].e);
        printf("Test case %d: ", ++cn);
        solve();
    }
    return 0;
}