题解【洛谷P2003】平板

题面

由于本题中\(n\)很小，\(\Theta(n^2)\)的暴力也可以通过。

具体可参照洛谷题解区

#include 
#define itn int
#define gI gi

using namespace std;

inline int gi()
{
    int f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return f * x;
}

int n, m, ans, cnt1, cnt2;
struct Node
{
    int y, x1, x2;
} a[103];

int main()
{
    //freopen(".in", "r", stdin);
    //freopen(".out", "w", stdout);
    n = gi();
    for (int i = 1; i <= n; i+=1) a[i].y = gi(), a[i].x1 = gi(), a[i].x2 = gi();
    for (int i = 1; i <= n; i+=1)
    {
        cnt1 = cnt2 = 0;
        for (int j = 1; j <= n; j+=1)
        {
            if (i == j || a[j].y >= a[i].y) continue;
            if (a[i].x1 < a[j].x2 && a[i].x1 >= a[j].x1) cnt1 = max(cnt1, a[j].y);
            if (a[i].x2 <= a[j].x2 && a[i].x2 > a[j].x1) cnt2 = max(cnt2, a[j].y);
        }
        ans += a[i].y * 2 - cnt1 - cnt2;
    }
    printf("%d\n", ans);
    return 0;
}