IME Starters Try-outs 2018 C. China Adventures

C. China Adventures

time limit per test

1.0 s

memory limit per test

512 MB

input

standard input

output

standard output

The ACM ICPC World Finals is a huge event! On 2018 the World Finals was at Beijing, China.

The chinese culture is way different than brazilian culture. They eat

(as they translated: "Bad fish chips")

and

(as they translated: "Sauteed chicken on the left")

Their traffic is crazy, with cars driving on the cycle path, motorcycles driving the wrong way or on the sidewalk. But, the weirdest of all, they use QR-codes for everything.

A QR-code is a square image, in this case a $\mathrm{21\times 2121\times 21 pixel\; matrix,\; that\; each\; pixel\; is\; either\; black\; or\; white.\; The\; code\; carries\; some\; information,\; usually\; an\; URL\; or\; a\; text,\; and\; can\; be\; read\; by\; almost\; all\; modern\; smartphones.}$

They use QR-codes so that people don't have to copy text all the time. They just have to open their smartphone's camera and the magic happens.

During a tour on Beijing, Lucas Santana, the Handsome, noticed that to visit some tourist places he needed to know the QR-code to buy the tickets for that place. Since he's a very smart guy, he decided to create a mobile application to draw the QR-codes, instead of searching for them.

The mobile application works as follow: it starts with a $\mathrm{21\times 2121\times 21 white\; pixel\; matrix.\; You\; can\; click\; any\; pixel\; to\; swap\; it\text{'}s\; color\; from\; white\; to\; black\; or\; from\; black\; to\; white\; in\; 1\; decisecond.\; Also,\; you\; can\; restart\; the\; application\; at\; any\; time\; to\; have\; an\; all\; white\; pixel\; matrix\; again,\; but\; it\; take aa deciseconds\; to\; do\; so.\; When\; you\; complete\; the\; QR-code,\; you\; can\; assume\; you\; can\; enter\; the\; tourist\; place.}$

Lucas knows the QR-code of $\mathrm{nn places\; he\; wants\; to\; visit.\; He\; doesn\text{'}t\; care\; about\; the\; order\; he\; visits\; the\; places,\; but,\; since\; he\; only\; have\; a\; few\; days\; left\; on\; China,\; he\; wants\; to\; spend\; the\; least\; time\; as\; possible\; drawing\; the\; QR-codes.\; Can\; you\; help\; him\; saying\; the\; minimum\; amount\; of\; deciseconds\; to\; draw\; all nn QR-codes,\; the\; order\; he\; should\; visit\; the\; places\; and\; when\; he\; should\; restart\; the\; application?}$

He will not visit the same place more than once. (Don't even try this. He will give you a wrong answer!).

Input

The first line of input contains two integers, $\mathrm{nn and aa (1\le n\le 181\le n\le 18, 1\le a\le 501\le a\le 50)\; —\; the\; number\; of\; places\; he\; wants\; to\; visit\; and\; the\; total\; time\; to\; restart\; the\; application.}$

The next lines contains the QR-codes of the places separated by empty lines. There will be $\mathrm{nn QR-codes,\; each\; having 2121 lines\; of 2121 digits\; being 00 or 11.\; Digit 00 means\; a\; white\; pixel\; and 11 a\; black\; pixel.}$

Output

Print the minimum amount of deciseconds to draw all QR-codes followed by at least $\mathrm{nn lines\; and\; at\; most 2\times n-12\times n-1 lines.\; Each\; line\; should\; have\; one\; number\; between 11 and nn,\; meaning\; that\; he\; must\; visit\; the\; place ii,\; or\; one\; character\; "*"\; (without\; quotes),\; meaning\; that\; he\; must\; restart\; the\; application.}$

All numbers should be distinct.

If there are multiple answers, print any of them.

Example

input

Copy

1 10
111111100101101111111
100000101101001000001
101110101100101011101
101110100101001011101
101110101000101011101
100000101001101000001
111111101010101111111
000000001111100000000
110100110110001110110
111101000100000110110
111110111110101101101
001110001011001101001
000111110010110100100
000000001001000100000
111111101110011110110
100000100011110001000
101110100101011100010
101110101111000010011
101110100100100110101
100000101010011100000
111111101101100010010

output

Copy

228
1

Note

In sample case, he should just draw the only QR-code available and visit the place. The QR-code is:

思路：求出转移代价相当求一个哈密顿路，状压dp即可。

#include <iostream>
#include <fstream>
#include <sstream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <list>
#include <iomanip>
#include <cctype>
#include <cassert>
#include <bitset>
#include <ctime>

using namespace std;

#define pau system("pause")
#define ll long long
#define pii pair<int, int>
#define pb push_back
#define mp make_pair
#define clr(a, x) memset(a, x, sizeof(a))

const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;

/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update> T;
*/

int n, a;
char s[22][22][22];
vector<int> e[22];
struct Edge {
    int v, w;
    Edge () {}
    Edge (int v, int w) : v(v), w(w) {}
};
int cal(char a[22][22], char b[22][22]) {
    int res = 0;
    for (int i = 1; i <= 21; ++i) {
        for (int j = 1; j <= 21; ++j) {
            res += a[i][j] != b[i][j];
        }
    }
    return res;
}
int cal2(char a[22][22]) {
    int dis = 0;
    for (int i = 1; i <= 21; ++i) {
        for (int j = 1; j <= 21; ++j) {
            dis += a[i][j] - '0';
        }
    }
    return dis;
}
int dp[1 << 20][20], dis[22][22];
pii pre[1 << 20][20];
int main() {
    scanf("%d%d", &n, &a);
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= 21; ++j) {
            scanf("%s", s[i][j] + 1);
        }
    }
    for (int i = 1; i <= n; ++i) {
        for (int j = i + 1; j <= n; ++j) {
            int d = cal(s[i], s[j]);
            dis[i][j] = dis[j][i] = d;
        }
    }
    for (int i = 1; i <= n; ++i) {
        int d = cal2(s[i]);
        dis[i][0] = dis[0][i] = d;
    }
    clr(dp, INF);
    dp[0][0] = 0;
    for (int i = 0; i < 1 << n; ++i) {
        for (int j = n; ~j; --j) {
            for (int k = 0; k <= n; ++k) {
                if (!k) {
                    if (dp[i][k] > dp[i][j] + a) {
                        dp[i][k] = dp[i][j] + a;
                        pre[i][k] = pii(i, j);
                    }
                }
                else {
                    if (i & (1 << k - 1)) continue;
                    int x = i | (1 << k - 1);
                    if (dp[x][k] > dp[i][j] + dis[j][k]) {
                        dp[x][k] = dp[i][j] + dis[j][k];
                        pre[x][k] = pii(i, j);
                    }
                }
            }
        }
    }
    int ans = INF;
    pii pp;
    for (int i = 0; i <= n; ++i) {
        if (ans > dp[(1 << n) - 1][i]) {
            ans = dp[(1 << n) - 1][i];
            pp = pii((1 << n) - 1, i);
        }
    }
    printf("%d
", ans);
    stack<pii> sta;
    while (pp != pii(0, 0)) {
        sta.push(pp);
        int x = pp.first, y = pp.second;
        pp = pre[x][y];
    }
    while (sta.size()) {
        pp = sta.top(); sta.pop();
        if (pp.second) {
            printf("%d
", pp.second);
        } else {
            puts("*");
        }
    }
    return 0;
}