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LOJ3156

题面就不放了 , 放一下数据范围 .

看到 \(n<=2000,m<=4000\) 就想到直接 \(dfs\) 到底 , 居然就过了前 \(4\)个 样例 , 最后一个要 \(2s\) . 后来写了 \(A=B=0\) 的 \(5\) 分 , 我知道写的是错的 , 还是交了以下这份代码 . ( LOJ 数据应该是官方数据 ) 得分 \(70\) .

晚上到 LOJ 上一测 , 发现如果直接跑我的暴力有 \(80\) 分 , 又到 \(AC\) 记录里面随便看了一篇比较优秀的 , 改在后面 .

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cassert>
#include<queue>
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define Debug(x) cout<<#x<<"="<<x<<endl
using namespace std;
typedef long long LL;
const int INF=1e9+7;
inline LL read(){register LL x=0,f=1;register char c=getchar();while(c<48||c>57){if(c=='-')f=-1;c=getchar();}while(c>=48&&c<=57)x=(x<<3)+(x<<1)+(c&15),c=getchar();return f*x;
}const int N=100005;
const int M=200005;struct Edge{int v,s,t,nxt;
}e[M];
int first[N],Ecnt=0;
inline void Add_edge(int u,int v,int s,int t){e[++Ecnt]=(Edge){v,s,t,first[u]};first[u]=Ecnt;
}int n, m, A, B, C;inline LL calc(int x){return 1ll * A * x * x + 1ll * B * x + C;
}namespace baoli{LL ans = INF;inline void dfs(int u, int time, LL cost){if(u == n){ans = min(ans, cost + time);}for(int i = first[u]; i; i = e[i].nxt){int v = e[i].v, s = e[i].s, t = e[i].t;if(s < time) continue;dfs(v, t, cost + calc(s - time));}}inline void main(){dfs(1, 0, 0);printf("%lld\n", ans);exit(0);}
};namespace Subtask1{ // A == 0 && B == 0int dis[N], time[N];queue <int> q;LL ans = INF;inline void Return(LL ans){printf("%lld\n", ans);exit(0);}inline void main(){q.push(1);memset(dis, 0x3f, sizeof dis);dis[1] = 0;while(!q.empty()){int u = q.front(); q.pop();for(int i = first[u]; i; i = e[i].nxt){int v = e[i].v;if(time[u] > e[i].s) continue;if(dis[u] + 1 < dis[v]){dis[v] = dis[u] + 1;time[v] = e[i].t;q.push(v);}if(v == n) ans = min(ans, 1ll * (dis[u] + 1 + 1) * C + e[i].t);}}Return(ans);assert(false);}/*inline void main(){q.push((Node){1, 0, 0});while(!q.empty()){int u = q.front().x, d = q.front().dis, t = q.front().time; q.pop();for(int i = first[u]; i; i = e[i].nxt){int v = e[i].v;if(e[i].t < t) continue;if(}}}*/
};int main(){
#ifndef filefreopen("route.in","r",stdin);freopen("route.out","w",stdout);
#endifn = read(), m = read(), A = read(), B = read(), C = read();for(register int i = 1; i <= m; ++i){register int x = read(), y = read(), p = read(), q = read();Add_edge(x, y, p, q);}if(n <= 2000 && m <= 4000) baoli::main();if(A == 0 && B == 0) Subtask1::main();
}

LOJ上的一份AC代码

我没注意到时间 \(q<=1000\) , 这样的话 \(O(nq)=O(1e8)\) 应该可以卡过?
直接 \(dp\) 有人得了 95分 . 把列车 按照时间排序 , 再依次更新 , 这样 \(1e8\) 就跑不满 , 直接就过了 . 可以用 \(vector\) 存状态 .
对于这个数据范围就当 \(O(nq)\) 是正解好了 .

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define Debug(x) cout<<#x<<"="<<x<<endl
using namespace std;
typedef long long LL;
const LL INF=1e18+7;
inline LL read(){register LL x=0,f=1;register char c=getchar();while(c<48||c>57){if(c=='-')f=-1;c=getchar();}while(c>=48&&c<=57)x=(x<<3)+(x<<1)+(c&15),c=getchar();return f*x;
}const int N = 1e5 + 5;
const int M = 2e5 + 5;struct Node{int x, y, p, q;
}a[M];
inline bool cmp1(Node a, Node b){if(a.p == b.p) return a.q < b.q;return a.p < b.p;
}vector <LL> f[N];
vector <int> t[N];
int n, m, A, B, C;inline LL calc(int x){return 1ll * A * x * x + 1ll * B * x + C;
}int main(){
#ifndef filefreopen("route.in","r",stdin);freopen("route.out","w",stdout);
#endifn = read(), m = read(), A = read(), B = read(), C = read();for(int i = 1; i <= m; ++i){a[i].x = read(), a[i].y = read(), a[i].p = read(), a[i].q = read();}sort(a + 1, a + m + 1, cmp1);f[1].push_back(0), t[1].push_back(0);for(int i = 1; i <= m; ++i){int x = a[i].x, y = a[i].y, p = a[i].p, q = a[i].q;int tt = -1;for(int j = 0; j < t[y].size(); ++j)if(t[y][j] == q) {tt = j; break;}for(int j = 0; j < f[x].size(); ++j){if(t[x][j] > p) continue;int len = p - t[x][j];if(tt == -1){f[y].push_back(f[x][j] + calc(len));t[y].push_back(q);tt = f[y].size() - 1;}else if(f[x][j] + calc(len) < f[y][tt]){f[y][tt] = f[x][j] + calc(len);}}}LL ans = INF;for(int i = 0; i < f[n].size(); ++i)ans = min(ans, f[n][i] + t[n][i]);printf("%lld\n", ans);
}