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POJ3067：Japan(线段树)

2024-09-19 21:19:07   215人阅读

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


求交叉的点数
满足交叉的条件是si<sj&&ei>ej || si>sj&&ei<ej
排好序后就是求终点逆序数了。能够用线段树实现


#include<stdio.h>
#include<string.h>
#include <algorithm>
using namespace std;

#define N 1010
#define M 1000010
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
int T,n,m,k;
int sum[N<<2],a[N];
__int64 ans;

struct node
{
    int x;
    int y;
} s[M];

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

void Pushup(int rt)
{
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}

void Update(int rt,int l,int r,int x)
{
    if(l == r)
    {
        sum[rt]++;
        a[l]++;
        return ;
    }
    int mid = (l + r) >> 1;
    if(x <= mid)
        Update(lson,x);
    else
        Update(rson,x);
    Pushup(rt);
}

int Query(int rt,int l,int r,int L,int R)
{
    if(L <= l && R >= r)
    {
        return sum[rt];
    }
    int mid= (l + r) >> 1;
    int res= 0;
    if(L <= mid) res += Query(lson,L,R);
    if(R > mid ) res += Query(rson,L,R);
    return res;
}

int main()
{
    int i,j,res,cas = 1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&k);
        memset(sum,0,sizeof(sum));
        memset(a,0,sizeof(a));
        ans = 0;
        res = 0;
        for(i = 1; i <= k; ++i)
            scanf("%d %d",&s[i].x,&s[i].y);
        sort(s+1,s+1+k,cmp);
        for(i = 1; i <= k; ++i)
        {
            int tmp = s[i].y;
            if(i>1 && s[i].y == s[i-1].y)
                res++;
            else
                res = 0;
            ans += Query(1,1,n,1,s[i].x)-a[s[i].x]-res;
            Update(1,1,n,s[i].x);
        }
        printf("Test case %d: %I64d\n",cas++,ans);
    }
    return 0;
}


POJ3067：Japan(线段树)


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