AtCoder Beginner Contest 286——E - Souvenir

AtCoder Beginner Contest 286 题目讲解

A题 B题 C题 D题 E题

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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 $N$ cities. There are also one-way direct flights that connect different cities.
The availability of direct flights is represented by $N$ strings $S_1,S_2,\dots ,S_N$​ of length $N$ each. If the $j$-th character of $S_i$​ is $Y$, there is a direct flight from city $i$ to city $j$; if it is $N$, there is not.
Each city sells a souvenir; city i sells a souvenir of value $A_i$​.

Consider the following problem:

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

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

Determine if he can travel from city $S$ to city $T$ 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 $Q$ pairs $(U_i,V_i)$ of distinct cities.
For each $1\le i\le Q$, print the answer to the problem above when $S=U_i$​ and $T=V_i$​.

Constraints

• $2\le N\le 300$
• $1\le A_i\le 10^9$
• $S_i$​ is a string of length
• $N$ consisting of $Y$ and $N$.
• The $i$-th character of $S_i$​ is $N$.
• $1\le Q\le N(N-1)$
• $1\le U_i,V_i\le N$
• $Ui\ne Vi$​
• If $i\ne j$, then $(U_i,V_i)\ne (U_j,V_J)$.
• $N,A_i,Q,U_i,$ and $V_i$​ 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 $Q$ lines.
The $i$-th ($1\le i\le Q$) line should contain Impossible if he cannot travel from city $U_i$​ to city $V_i$​; 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 $(S,T)=(U_1,V_1)=(1,3)$, there is a direct flight from city $1$ to city $3$, so the minimum possible number of direct flights is $1$, which is achieved when he uses that direct flight. In this case, the total value of the souvenirs is $A_1+A_3=30+70=100$.
For $(S,T)=(U2,V2)=(3,1)$, the minimum possible number of direct flights is $2$. The following two routes achieve the minimum: cities $3\to 4\to 1$, and cities $3\to 5\to 1$. Since the total value of souvenirs in the two routes are $70+20+30=120$ and $70+60+30=160$, respectively, so he chooses the latter route, and the total value of the souvenirs is $160$.
For $(S,T)=(U_3,V_3)=(4,5)$, the number of direct flights is minimum when he travels cities $4\to 1\to 3\to 5$, where the total value of the souvenirs is $20+30+70+60=180$.

Sample Input 2

2
100 100
NN
NN
1
1 2

Sample Output 2

Impossible

There may be no direct flight at all.

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思路

这道题一看就可以用我们的$Dijkstra$了，但是，$Dijkstra$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$res$，记录价值。

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

不过每次调用$Dijkstra$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$O(N^3)$的时间复杂度，就可以过了！

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代码

#include 
#include 
#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;
}

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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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