Codeforces 853B Jury Meeting

题意

从城市1-n来的评审团到城市0商讨国家大事，离开和抵达的那一天不能讨论，飞机均当天抵达，给出所有飞机起飞抵达代价情况，问能否使所有评审员聚齐连续k天并返回，并求最小代价

思路

从前向后扫一遍，求每天的出发最小代价L[i]，从后向前扫，求每天最小离开代价R[i]

从前向后扫一遍，每天的最小代价为L[i]+R[i+k+1]

将每天的默认大小设为1e12，因为最大代价不超过1e11，可以据此确定答案是否合法

代码

#include<bits/stdc++.h>
using namespace std;
int n,m,k,d,f,t,c;
typedef long long ll;
const ll inf = 1e12;
const int maxn = 1e6+;
ll l[maxn], r[maxn], minn[maxn];
vector< pair<int, int> >to[maxn], back[maxn];
int main(){
    scanf("%d %d %d",&n, &m, &k);
    for(int i = ;i<m;i++){
        scanf("%d %d %d %d",&d, &f,&t, &c);
        if(t == ) to[d].push_back({f,c});
        else back[d].push_back({t,c});
    }
    ll best = inf*n;
    for(int i = ;i<maxn;i++) minn[i] = inf;
    for(int i = ;i<maxn;i++){
        for(pair<int,int> p : to[i]){
            int dest = p.first, cost = p.second;
            if(cost<minn[dest]) best-= (minn[dest]-cost), minn[dest] = cost;
        }
        l[i] = best;
    }
    best = inf*n;
    for(int i = ;i<maxn;i++) minn[i] = inf;
    for(int i = maxn-;i>=;i--){
        for(pair<int,int> p : back[i]){
            int dest = p.first, cost = p.second;
            if(cost<minn[dest]) best-=(minn[dest]-cost), minn[dest] = cost;
        }
        r[i] = best;
    }
    ll ans = inf*n;
    for(int i = ;i<maxn-k-;i++){
        int rr = i+k+;
        ans = min(ans,l[i]+r[rr]);
    }
    if(ans>=inf) return *printf("-1");
    return *printf("%I64d",ans);
}