接上一篇文章;
这里直接把左端点和右端点映射到vector数组上;
映射一个open和close数组;
枚举1..2e5
如果open[i]内有安排;
则用那个安排和dp数组来更新答案;
更新答案完之后,如果有close数组
则把close数组里面的安排用来更新dp数组;
#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)
#define Open() freopen("F:\rush.txt","r",stdin)
#define Close() ios::sync_with_stdio(0)
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 2e5;
const int INF = 2e9+10;
struct abc{
int l,r,cost;
};
int n,x,dp[N],ans = INF;
vector <abc> open[N+100],close[N+100];
int main(){
//Open();
Close();
cin >> n >> x;
abc temp;
rep1(i,1,n){
cin >> temp.l >> temp.r >> temp.cost;
open[temp.l].pb(temp);
close[temp.r].pb(temp);
}
int len;
rep1(i,1,N){
while (!open[i].empty()){
temp = open[i].back();
open[i].pop_back();
len = temp.r-temp.l+1;
if (x>=len){
if (dp[x-len] > 0)
ans = min(ans,dp[x-len]+temp.cost);
}
}
while (!close[i].empty()){
temp = close[i].back();
close[i].pop_back();
len = temp.r-temp.l+1;
if (dp[len]==0)
dp[len] = temp.cost;
else
dp[len] = min(dp[len],temp.cost);
}
}
if (ans==INF)
cout << -1 << endl;
else
cout << ans << endl;
return 0;
}