codeforces 854 D. Jury Meeting(dp水题)

题目链接

D. Jury Meeting

分析

昨晚打正式赛的时候太急了，代码写的有点毒，下了课后一看，立马发现bug…..

将去的航班表和返回航班表分别按照时间最小和最大排序，对于去的航班来说，挨个扫描过去，对每一个人维护当前的最小花费，若时间总的航班满足等于　n 说明这一天可以作为起始时间，更新这一天的花费dp[d]=now_ans,now_ans 记录当前最小花费，这个更新是O(1) d的具体看代码.
对与返程航班也这样干就行了，计算出可以作为返回时间点的最小花费dp1.
对两个dp数组分别求到目前为止的最小花费,如dp[i]=min(dp[i],dp[i−1]) 前i天出发的最小花费，反向数组亦然.

ac code

#include<bits/stdc++.h>#define pb push_back#define mp make_pair#define PI acos(-1)#define fi first#define se second#define INF 0x3f3f3f3f#define INF64 0x3f3f3f3f3f3f3f3f#define random(a,b) ((a)+rand()%((b)-(a)+1))#define ms(x,v) memset((x),(v),sizeof(x))#define scint(x) scanf("%d",&x );#define scf(x) scanf("%lf",&x );#define eps 1e-10#define dcmp(x) (fabs(x) < eps? 0:((x) <0?-1:1))#define lc o<<1#define rc o<<1|1using namespace std;typedef long long LL;typedef long double DB;typedef pair<int,int> Pair;const int maxn = 1e6+10;const int MAX_V= 500+10;const int MOD = 998244353;struct day{    int d,cost,num;    bool operator< (const day& o)const{        if(d != o.d)return d > o.d;        else return cost > o.cost;    }};day P[maxn],b[maxn];bool cmp(const day & a,const day & b){    if(a.d != b.d)return a.d >b.d;    else return a.cost < b.cost;}int c[maxn],d[maxn];LL dp[maxn],dp1[maxn];int main(){    ios_base::sync_with_stdio(0);    cin.tie(0);    cout.tie(0);    int n,m,k;    cin>>n>>m>>k;    int t1 =0;    int t2 =0;    for(int i=1 ; i<maxn ; ++i) dp[i] = INF64,c[i] = 0,dp1[i] = INF64,d[i]=0;    priority_queue<day> Q;    for(int i=1 ; i<=m ; ++i){        int d,u,v,cost;cin>>d>>u>>v>>cost;        if(u)Q.push(day{d,cost,u});        else P[t2++] = day{d,cost,v};    }    sort(P,P+t2,cmp);    int tot =0;    LL ans =0;    while (!Q.empty()) {        day p = Q.top();Q.pop();        int u = p.num;        if(!c[u])c[u] = p.cost,tot++,ans += c[u];        else if( c[u]> p.cost) ans += p.cost - c[u],c[u] = p.cost;        if(tot == n)dp[p.d] = min(dp[p.d],ans);    }    tot =0;    ans =0;    for(int i = 0 ; i<t2 ; ++i){        day p  = P[i];        int v = p.num;        if(!d[v])d[v] = p.cost,tot++,ans += d[v];        else if(d[v] > p.cost) ans += p.cost - d[v],d[v] = p.cost;        if(tot == n)dp1[p.d] = min(dp1[p.d],ans);    }    for(int i = 2 ; i<maxn ; ++i)dp[i] = min(dp[i],dp[i-1]);    //std::cout << maxn << '\n';    for(int i = maxn -2 ; i>=0 ; --i){        dp1[i] = min(dp1[i],dp1[i+1]);    }    ans = INF64;    for(int i=1 ; i<maxn -k-1; ++i){        ans = min(ans , dp[i]+ dp1[i+k+1]);    }    // for(int i=1 ; i<20 ; ++i){    //     std::cout << dp[i]<<" " << dp1[i] << '\n';    // }    if(ans == INF64)std::cout << -1 << '\n';    else std::cout << ans << '\n';    //std::cout << "time "<< clock()/1000 <<"ms"<< '\n';    return 0;}