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  • 【Codeforces Round #422 (Div. 2) C】Hacker, pack your bags!(hash写法)


    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;
    }
    
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  • 原文地址:https://www.cnblogs.com/AWCXV/p/7626221.html
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