AtCoder Beginner Contest 286 题目讲解

A题 B题 C题 D题 E题

蒟蒻来讲题，还望大家喜。若哪有问题，大家尽可提！

Hello, 大家好哇！本初中生蒟蒻今天讲解一下AtCoder Beginner Contest 286的E题——Souvenir

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原题

Problem Statement

There are N N N cities. There are also one-way direct flights that connect different cities.
The availability of direct flights is represented by N N N strings S 1 , S 2 , … , S N S_{1},S_{2},…,S_{N} S1,S2,…,SN of length N N N each. If the j j j-th character of S i S_{i} Si is Y Y Y, there is a direct flight from city i i i to city j j j; if it is N N N, there is not.
Each city sells a souvenir; city i sells a souvenir of value A i A_{i} Ai.

Consider the following problem:

Takahashi is currently at city S S S and wants to get to city T T T (that is different from city S S S) using some direct flights.
Every time he visits a city (including S S S and T T T), he buys a souvenir there.
If there are multiple routes from city S S S to city T T T, Takahashi decides the route as follows:

He tries to minimize the number of direct flights in the route from city S to city T T T.
Then he tries to maximize the total value of the souvenirs he buys.

Determine if he can travel from city S S S to city T T T using the direct flights. If he can, find the “number of direct flights” and “total value of souvenirs” in the route that satisfies the conditions above.

You are given Q Q Q pairs ( U i , V i ) (U_{i},V_{i}) (Ui,Vi) of distinct cities.
For each 1 ≤ i ≤ Q 1\leq i\leq Q 1≤i≤Q, print the answer to the problem above when S = U i S=U_{i} S=Ui and T = V i T=V_{i} T=Vi.

Constraints

2 ≤ N ≤ 300 2\leq N\leq 300 2≤N≤300
1 ≤ A i ≤ 1 0 9 1\leq A_{i}\leq10^9 1≤Ai≤109
S i S_{i} Si is a string of length
N N N consisting of Y Y Y and N N N.
The i i i-th character of S i S_{i} Si is N N N.
1 ≤ Q ≤ N ( N − 1 ) 1\leq Q\leq N(N−1) 1≤Q≤N(N−1)
1 ≤ U i , V i ≤ N 1\leq U_{i},V_{i}\leq N 1≤Ui,Vi≤N
U i ≠ V i Ui\neq Vi Ui=Vi
If i ≠ j i\neq j i=j, then ( U i , V i ) ≠ ( U j , V J ) (U_{i},V_{i})\neq (U_{j},V_{J}) (Ui,Vi)=(Uj,VJ).
N , A i , Q , U i , N,A_{i},Q,U_{i}, N,Ai,Q,Ui, and V i V_{i} Vi are all integers.

Input

The input is given from Standard Input in the following format:

N
A1  A2  …  AN
S1
S2
⋮
SN
Q
U1  V1
U2  V2
⋮
UQ  VQ

Output

Print Q Q Q lines.
The i i i-th ( 1 ≤ i ≤ Q 1\leq i\leq Q 1≤i≤Q) line should contain Impossible if he cannot travel from city U i U_{i} Ui to city V i V_{i} Vi; if he can, the line should contain “the number of direct flights” and “the total value of souvenirs” in the route chosen as described above, in this order, separated by a space.

Sample Input 1

5
30 50 70 20 60
NYYNN
NNYNN
NNNYY
YNNNN
YNNNN
3
1 3
3 1
4 5

Sample Output 1

1 100
2 160
3 180

For ( S , T ) = ( U 1 , V 1 ) = ( 1 , 3 ) (S,T)=(U_{1},V_{1})=(1,3) (S,T)=(U1,V1)=(1,3), there is a direct flight from city 1 1 1 to city 3 3 3, so the minimum possible number of direct flights is 1 1 1, which is achieved when he uses that direct flight. In this case, the total value of the souvenirs is A 1 + A 3 = 30 + 70 = 100 A_{1}+A_{3}=30+70=100 A1+A3=30+70=100.
For ( S , T ) = ( U 2 , V 2 ) = ( 3 , 1 ) (S,T)=(U2,V2)=(3,1) (S,T)=(U2,V2)=(3,1), the minimum possible number of direct flights is 2 2 2. The following two routes achieve the minimum: cities 3 → 4 → 1 3→4→1 3→4→1, and cities 3 → 5 → 1 3→5→1 3→5→1. Since the total value of souvenirs in the two routes are 70 + 20 + 30 = 120 70+20+30=120 70+20+30=120 and 70 + 60 + 30 = 160 70+60+30=160 70+60+30=160, respectively, so he chooses the latter route, and the total value of the souvenirs is 160 160 160.
For ( S , T ) = ( U 3 , V 3 ) = ( 4 , 5 ) (S,T)=(U_{3},V_{3})=(4,5) (S,T)=(U3,V3)=(4,5), the number of direct flights is minimum when he travels cities 4 → 1 → 3 → 5 4→1→3→5 4→1→3→5, where the total value of the souvenirs is 20 + 30 + 70 + 60 = 180 20+30+70+60=180 20+30+70+60=180.

Sample Input 2

Sample Output 2

Impossible

There may be no direct flight at all.

思路

这道题一看就可以用我们的 D i j k s t r a Dijkstra Dijkstra了，但是， D i j k s t r a Dijkstra Dijkstra只能求出最短路径，不能求出最短路径的最大价值。不过，我们可以再开一个数组 r e s res res，记录价值。

如果 d i s t [ i ] [ j ] > d i s t [ i ] [ t ] + g [ t ] [ j ] dist[i][j] > dist[i][t]+g[t][j] dist[i][j]>dist[i][t]+g[t][j]就说明有一条最短路径，因为短优先，所以直接让 r e s [ i ] [ j ] = r e s [ i ] [ t ] + a [ j ] res[i][j] = res[i][t] + a[j] res[i][j]=res[i][t]+a[j]
如果 d i s t [ i ] [ j ] = d i s t [ i ] [ t ] + g [ t ] [ j ] dist[i][j] = dist[i][t]+g[t][j] dist[i][j]=dist[i][t]+g[t][j]就说明最短路径相等，那我们就要求最大值，即为 r e s [ i ] [ j ] = m a x ( r e s [ i ] [ j ] , r e s [ i ] [ t ] + a [ j ] ) res[i][j] = max(res[i][j], res[i][t]+a[j]) res[i][j]=max(res[i][j],res[i][t]+a[j])

不过每次调用 D i j k s t r a Dijkstra Dijkstra会超时，所以我们要先预处理一下，把任一点到任一点枚举一遍因为点的数量很少，这样就是 O ( N 3 ) O(N^3) O(N3)的时间复杂度，就可以过了！

代码

#include <iostream>
#include <cstring>
#define int long long //本题要开long long！using namespace std;const int N = 5e2 + 10;int n, m;
int g[N][N], dist[N][N]; //定义邻接矩阵和最短距离数组
int cost[N];
int res[N][N];//最短距离下的最大价值
bool st[N];void dijkstra()
{for (int i = 0; i <= n; i ++)for (int j = 0; j <= n; j ++)dist[i][j] = 1e18;for (int i = 1; i <= n; i ++){dist[i][i] = 0;memset(st, 0, sizeof st);for (int k = 0; k < n - 1; k ++){int t = -1;for (int j = 1; j <= n; j ++)if (!st[j] && (t == -1 || dist[i][t] > dist[i][j]))t = j;st[t] = 1;for (int j = 1; j <= n; j ++){if (dist[i][j] > dist[i][t] + g[t][j]) //说明有最短路径{dist[i][j] = dist[i][t] + g[t][j];res[i][j] = res[i][t] + cost[j]; //必须选}else if (dist[i][j] == dist[i][t] + g[t][j]) //长度相等res[i][j] = max(res[i][j], res[i][t] + cost[j]); //选择价值大的}}}
}signed main()
{cin >> n;for (int i = 1; i <= n; i ++)for (int j = 1; j <= n; j ++)g[i][j] = 1e18;for (int i = 1; i <= n; i ++)cin >> cost[i];for (int i = 1; i <= n; i ++){string s;cin >> s;for (int j = 0; j < n; j ++)if (s[j] == 'Y')g[i][j + 1] = 1; //边权赋值为1}dijkstra();cin >> m;int a, b;while (m --){cin >> a >> b;if (dist[a][b] >= 1e18) cout << "Impossible" << endl;else cout << dist[a][b] << " " << res[a][b] + cost[a] << endl; //要加起点！}return 0;
}

今天就到这里了！

大家有什么问题尽管提，我都会尽力回答的！最后，除夕夜祝大家新年快乐！

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AtCoder Beginner Contest 286——E - Souvenir

AtCoder Beginner Contest 286 题目讲解

原题

思路

代码

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