POJ3067：Japan(线段树)

Description

Input

Output

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，可以用线段树实现

#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),布布扣,bubuko.com

POJ3067：Japan(线段树)

原文：http://blog.csdn.net/libin56842/article/details/25664691