題意：給你兩個(gè)函數(shù)水平范圍是[0,10]，定義第三個(gè)函數(shù)是該區(qū)間上兩個(gè)函數(shù)的最大值。水平線就是第三個(gè)函數(shù)的最小值。
解法：排序模擬就好。
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#define N 1005
inline double max(double a, double b)
{
return a > b ? a : b;
}
struct segement
{
double height;
double start, end;
}a[N], b[N];
double step[N];
int main()
{
int n1, n2, n;
double sum;
while(~scanf("%d", &n1))
{
n = 1;
sum = 0;
for(int i = 0; i < n1; i++)
{
scanf("%lf %lf", &a[i].height, &a[i].end);
a[i].start = sum;
sum += a[i].end;
a[i].end = sum;
step[n++] = sum;
}
scanf("%d", &n2);
sum = 0;
for(int i = 0; i < n2; i++)
{
scanf("%lf %lf", &b[i].height, &b[i].end);
b[i].start = sum;
sum += b[i].end;
b[i].end = sum;
step[n++] = sum;
}
std::sort(step, step + n);
double ans = 1e9, height1, height2;
for(int i = 1; i < n; i++)
{
for(int j = 0; j < n1; j++)
if(a[j].start <= step[i - 1] && a[j].end >= step[i])
{
height1 = a[j].height;
break;
}
for(int j = 0; j < n2; j++)
if(b[j].start <= step[i - 1] && b[j].end >= step[i])
{
height2 = b[j].height;
break;
}
if(max(height1, height2) < ans) ans = max(height1, height2);
}
printf("%.3lf\n", ans);
}
return 0;
}