Poj 3067 Japan【树状数组】

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Japan

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 25612		Accepted: 6930

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

题目大意：

问题可以抽象看成上边有一排点，下边有一排点（上边n个）（下边m个），一共要建K条边，问有多少个交叉点（注意，这里在节点处的交叉点不算）、

思路：

1、首先我们明确两条线段会有交叉点的情况：

①对于两条边（a，b），（c，d）需要有a>c&&b<d||a<c&&b>d，才会有交叉点出现。

②我门还能明确这样一个问题：线段的建设时间是不会影响最终结果的。

2、那么我们可以根据上述的第二条，我们对二元组进行排序，使得（x，y）中，x按照从大到小排序，y也从大到小排序，例如样例得到结果：

3 2

3 1

2 3

1 4

那么这个时候我们再按照这个顺序考虑这个问题，其实就不用管第一个数了，我们保证了递减的性质，我们只要找当前y元素之间有多少个比他小的就行了、问题转化到：

2 1 3 4 找逆序对个数。

3、直接暴力找会超时，我们使用树状数组来优化问题，然后剩下的就是实现代码的过程了。

4、注意在节点处的交叉点不算数，所以我们在查询的时候要查询（a【i】.y-1），来去除节点处的交叉点。另外数据会比较大，需要使用long long

Ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define ll long long int
struct node
{
    ll x,y;
}a[1000015];
ll n,m,q;
ll tree[1000015];//树
ll cmp(node a,node b)
{
    if(a.x==b.x)return a.y>b.y;
    else return a.x>b.x;
}
ll lowbit(ll x)//lowbit
{
    return x&(-x);
}
ll sum(ll x)//求和求的是比当前数小的数字之和，至于这里如何实现，很简单：ll sum=sum(a[i])；
{
    ll sum=0;
    while(x>0)
    {
        sum+=tree[x];
        x-=lowbit(x);
    }
    return sum;
}
void add(ll x,ll c)//加数据。
{
    while(x<=n)
    {
        tree[x]+=c;
        x+=lowbit(x);
    }
}
int main()
{
    ll t;
    ll kase=0;
    scanf("%I64d",&t);
    while(t--)
    {
        memset(a,0,sizeof(a));
        memset(tree,0,sizeof(tree));
        scanf("%lld%lld%lld",&m,&n,&q);
        for(ll i=0;i<q;i++)
        {
            scanf("%lld%lld",&a[i].x,&a[i].y);
        }
        sort(a,a+q,cmp);
       // ll i=0;
        ll output=0;
        for(int i=0;i<q;i++)
        {
            output+=sum(a[i].y-1);
            add(a[i].y,1);
        }
        printf("Test case %lld: %lld\n",++kase,output);
    }
}
/*
100
5 5 4
3 1
3 2
3 3
1 3
5 5 5
3 1
3 2
3 3
1 3
2 2
5 5 6
2 1
3 1
4 1
5 1
1 2
1 3
4 4 4
1 2
1 2
1 2
2 1
4 4 5
3 1
3 2
2 3
2 4
1 4
3 4 4
1 4
2 3
3 2
3 1
*/