[NOI2019]回家路线

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
", ans);
		exit(0);
	}
};

namespace Subtask1{ // A == 0 && B == 0
	int dis[N], time[N];
	queue <int> q;
	LL ans = INF;
	inline void Return(LL ans){
		printf("%lld
", 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 file
    freopen("route.in","r",stdin);
    freopen("route.out","w",stdout);
#endif
	n = 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 file
    freopen("route.in","r",stdin);
    freopen("route.out","w",stdout);
#endif
	n = 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
", ans);
}