Hacker, pack your bags!

題意：給n個區間，每個區間的長度為ri-li+1，權值為ci，選取2個不相交的區間，長度加起來為x，且權值和最小

思路：遍歷 l ，vi[l]存起點為l的區間的權值最小值，每次加入li==l的區間更新vi的值，更新後再更新ri==l-1的答案，這樣可以保證，每次選擇一個區間後，從vi裏選擇的另一個區間一定是不相交的

AC代碼：

#include<bits/stdc++.h>
#include "iostream"
#include "string.h"
#include "stack"
#include "queue"
#include 
 
"string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#define ll long long
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a) memset(a,0,sizeof(a))
#define mp(x,y) make_pair(x,y)
#define pb push_back
const long
 
 long INF = 1e18+1LL;
const int inf = 1e9+1e8;
using namespace std;
const int N=1e5+100;

vector<pair<ll,ll> > vexl[2*N],vexr[2*N];
ll n,x,vi[2*N],mk[2*N];
int main(){
    for(int i=0; i<=2e5; ++i) vi[i]=2LL*inf; //for(int i=1; i<=10; ++i) cout<<vi[i]<<" ";  
    ll ans=2LL*inf;
    scanf(
 
"%lld%lld",&n,&x);
    int li,ri,ma=0;
    ll ci;
    for(int i=1; i<=n; ++i){
        scanf("%d%d%lld",&li,&ri,&ci);
        ma=max(ma,ri);
        vexl[li].pb(mp(ri,ci));
        vexr[ri].pb(mp(li,ci));
    }
    for(int l=ma; l>0; --l){
        for(auto j :vexl[l]){
            int len=j.first-l+1;
            mk[len]=1,vi[len]=min(vi[len],j.second); //cout<<j.second<<endl;
        }
        int r=l-1;
        for(auto j :vexr[r]){
            int len=x-(r-j.first+1);
            if(len>0 && mk[len]){
                ans=min(ans,j.second+vi[len]); //cout<<len<<" "<<r<<" "<<j.second<<" "<<vi[len]<<endl;
            }
        }
    }
    if(ans==2LL*inf) printf("-1");
    else printf("%lld",ans);
    return 0;
}

Codeforces Round #422 C