POJ3067:Japan(树状数组求逆序对)

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.

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.

For each test case write one line on the standard output: 
Test case (case number): (number of crossings)

1
3 4 4
1 4
2 3
3 2
3 1

Test case 1: 5

Southeastern Europe 2006

题意：

两边分别有n,m个点，每边点的编号从上到下位1,2,3。。。

现在有K条边连接两边的点，最多只有两条边相交于一点而不会出现三条边以上交于一点的状况，问一共有几个交点

思路：

相交的情况必然是在在Li<Lj的情况下，Ri>Rj

所以在对L，R排序之后，只需要看R里面逆序数的多少即可

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <list>
#include <algorithm>
#include <climits>
using namespace std;

#define lson 2*i
#define rson 2*i+1
#define LS l,mid,lson
#define RS mid+1,r,rson
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N (1005*1005)
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define lowbit(x) (x&-x)
const int mod = 1e9+7;

struct node
{
    int l,r,id;
} a[N];
int n,c[N],x,y;

int cmp(node a,node b)
{
    if(a.l!=b.l)
        return a.l<b.l;
    return a.r<b.r;
}

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

void add(int x,int d)
{
    while(x<=y)
    {
        c[x]+=d;
        x+=lowbit(x);
    }
}

LL ans;
int main()
{
    int i,j,k,l,r,t,cas = 1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&x,&y,&n);
        MEM(c,0);
        for(i = 1; i<=n; i++)
        {
            scanf("%d%d",&a[i].l,&a[i].r);
        }
        sort(a+1,a+1+n,cmp);
        ans = 0;
        for(i = 1; i<=n; i++)
        {
            add(a[i].r,1);
            ans+=sum(y)-sum(a[i].r);
        }
        printf("Test case %d: %lld\n",cas++,ans);
    }

    return 0;
}