zoj 3574 求给定区间内n条线段能将区间内的区域划分成几个区域

答案是：n+1+逆序数的数量

因为如果n条线段都不相交，一定可以划分成n+1个区域，然后就画图YY一下

#include <cstdio>  
#include <algorithm>
using namespace std;
int left[50003], right[50003];  
int cnt;  
struct node {
    int y1,y2;
}p[30010];
int cmp(node a,node b){
    return a.y1<b.y1;
}
void merge(int a[], int l, int m, int r)  {  
    int i, j, k, n1, n2;  
    n1 = m - l + 1;  
    n2 = r - m;  
    for (i = 0; i < n1; i++)  
        left[i] = a[l+i];  
    for (i = 0; i < n2; i++)  
        right[i] = a[m+i+1];  
    left[n1] = right[n2] = 0x7fffffff;  
    i = j = 0;  
    for (k = l; k <= r; k++)  {  
        if (left[i] <= right[j])   a[k] = left[i++];  
        else  {  
            a[k] = right[j++];  
            cnt += n1 - i;   
        }  
    }  
}  
void mergeSort(int a[], int l, int r)  {  
    if (l < r)  {  
        int m = (l + r) / 2;  
        mergeSort(a, l, m);  
        mergeSort(a, m+1, r);  
        merge(a, l, m, r);  
    }  
} 
int  a[50011];
int main()  {  
    int n;  
    int l,r;
    while (scanf("%d%d",&l,&r)!=EOF)  {  
    scanf("%d",&n);
    int k,b;
    for(int i=0;i<n;i++){
        scanf("%d%d",&k,&b);
        p[i].y1=k*l+b;
        p[i].y2=k*r+b;
    }
    sort(p,p+n,cmp);
    for(int i=0;i<n;i++)        a[i]=p[i].y2;
     cnt=0;mergeSort(a, 0, n-1);  
     printf("%d\n", n+cnt+1);  
    }  
}