Codeforces Round #581 (Div. 2) C. Anna, Svyatoslav and Maps （Floyd 算法，最短路）

C. Anna, Svyatoslav and Maps

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

The main characters have been omitted to be short.

You are given a directed unweighted graph without loops with n vertexes and a path in it (that path is not necessary simple) given by a sequence p1,p2,…,pm of m vertexes; for each 1≤i<m there is an arc from pi to pi+1.

Define the sequence v1,v2,…,vk of k vertexes as good, if v is a subsequence of p, v1=p1, vk=pm, and p is one of the shortest paths passing through the vertexes v1, …, vk in that order.

A sequence a is a subsequence of a sequence b if a can be obtained from b by deletion of several (possibly, zero or all) elements. It is obvious that the sequence p is good but your task is to find the shortest good subsequence.

If there are multiple shortest good subsequences, output any of them.

Input

The first line contains a single integer n (2≤n≤100) — the number of vertexes in a graph.

The next n lines define the graph by an adjacency matrix: the j-th character in the i-st line is equal to 1 if there is an arc from vertex i to the vertex j else it is equal to 0. It is guaranteed that the graph doesn't contain loops.

The next line contains a single integer m (2≤m≤106) — the number of vertexes in the path.

The next line contains m integers p1,p2,…,pm (1≤pi≤n) — the sequence of vertexes in the path. It is guaranteed that for any 1≤i<m there is an arc from pi to pi+1.

Output

In the first line output a single integer k (2≤k≤m) — the length of the shortest good subsequence. In the second line output k integers v1, …, vk (1≤vi≤n) — the vertexes in the subsequence. If there are multiple shortest subsequences, print any. Any two consecutive numbers should be distinct.

Examples

inputCopy

4

0110

0010

0001

1000

4

1 2 3 4

outputCopy

3

1 2 4

inputCopy

4

0110

0010

1001

1000

20

1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4

outputCopy

11

1 2 4 2 4 2 4 2 4 2 4

inputCopy

3

011

101

110

7

1 2 3 1 3 2 1

outputCopy

7

1 2 3 1 3 2 1

inputCopy

4

0110

0001

0001

1000

3

1 2 4

outputCopy

2

1 4

Note

Below you can see the graph from the first example:

The given path is passing through vertexes 1, 2, 3, 4. The sequence 1−2−4 is good because it is the subsequence of the given path, its first and the last elements are equal to the first and the last elements of the given path respectively, and the shortest path passing through vertexes 1, 2 and 4 in that order is 1−2−3−4. Note that subsequences 1−4 and 1−3−4 aren't good because in both cases the shortest path passing through the vertexes of these sequences is 1−3−4.

In the third example, the graph is full so any sequence of vertexes in which any two consecutive elements are distinct defines a path consisting of the same number of vertexes.

In the fourth example, the paths 1−2−4 and 1−3−4 are the shortest paths passing through the vertexes 1 and 4.

题意：

给你了一个含有n个节点的有向图，

和一个序列p，

让你找一个最小的序列v，使其v[1]=p[1] ,v[end]=p[end] ，并且 v 中节点再遍历的过程中，p序列是最短路序列之一。

思路：

用Floyd 算法，nnn 算出任意两个的最短路径。

然后处理p序列，

以一个开始位st 向后找节点now 是否满足 now -st 满足 p[st] 到 p[now] 的最短路径距离。

如果满足就把now加入一个deque中待用，同时删除掉当前deque中已有的数，

否则就用deque中的数代替st，重复此操作。

细节见代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#include <stack>

#include <map>

#include <set>

#include <vector>

#include <iomanip>

#define ALL(x) (x).begin(), (x).end()

#define sz(a) int(a.size())

#define all(a) a.begin(), a.end()

#define rep(i,x,n) for(int i=x;i<n;i++)

#define repd(i,x,n) for(int i=x;i<=n;i++)

#define pii pair<int,int>

#define pll pair<long long ,long long>

#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)

#define MS0(X) memset((X), 0, sizeof((X)))

#define MSC0(X) memset((X), '\0', sizeof((X)))

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define eps 1e-6

#define gg(x) getInt(&x)

#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl

using namespace std;

typedef long long ll;

ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}

ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}

ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}

inline void getInt(int* p);

const int maxn = 1000010;

const int inf = 0x3f3f3f3f;

/*** TEMPLATE CODE * * STARTS HERE ***/

char s[105][105];

int n;

int m;

int a[maxn];

int cnt[105][105];

int dis[105][105];

int main()

{

    //freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);

    //freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);

    scanf("%d", &n);

    repd(i, 1, n)

    {

        scanf("%s", s[i] + 1);

    }

    scanf("%d", &m);

    repd(i, 1, m)

    {

        scanf("%d", &a[i]);

    }

    repd(i, 1, n)

    {

        repd(j, 1, n)

        {

            dis[i][j] = inf;

        }

        dis[i][i] = 0;

    }

    repd(i, 1, n)

    {

        repd(j, 1, n)

        {

//            cout<<s[i][j]<<" ";

            if (s[i][j] == '1')

            {

                dis[i][j] = 1;

            }

        }

//        cout<<endl;

    }

    repd(k, 1, n)

    {

        repd(i, 1, n)

        {

            repd(j, 1, n)

            {

                if (dis[i][k] + dis[k][j] < dis[i][j])

                {

                    dis[i][j] = dis[i][k] + dis[k][j];

                }

            }

        }

    }

//    repd(i,1,n)

//    {

//        repd(j,1,n)

//        {

//            cout<<dis[i][j]<<" ";

//        }

//        cout<<endl;

//    }

    deque<int> q;

    while (!q.empty())

    {

        q.pop_back();

    }

    std::vector<int> ans;

    ans.clear();

    int now = 2;

    int start = 1;

    while (now <= m)

    {

        int dist = now - start;

        if (dist == dis[a[start]][a[now]])

        {

            if (!q.empty())

            {

                q.pop_front();

            }

            q.push_back(now);

            now++;

        } else

        {

            ans.push_back(a[start]);

            if (!q.empty())

            {

                start = q.front();

                q.pop_front();

            }

        }

        // cout << sz(q) << endl;

    }

    ans.push_back(a[start]);

    if (ans[sz(ans) - 1] != a[m])

    {

        ans.push_back(a[m]);

    }

    cout << sz(ans) << endl;

    for (auto x : ans)

    {

        cout << x << " ";

    }

    cout << endl;

    return 0;

}

inline void getInt(int* p) {

    char ch;

    do {

        ch = getchar();

    } while (ch == ' ' || ch == '\n');

    if (ch == '-') {

        *p = -(getchar() - '0');

        while ((ch = getchar()) >= '0' && ch <= '9') {

            *p = *p * 10 - ch + '0';

        }

    }

    else {

        *p = ch - '0';

        while ((ch = getchar()) >= '0' && ch <= '9') {

            *p = *p * 10 + ch - '0';

        }

    }

}

巴特西

Codeforces Round #581 (Div. 2) C. Anna, Svyatoslav and Maps （Floyd 算法，最短路）

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