Codeforces 436C

题目链接

C. Dungeons and Candies

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

During the loading of the game "Dungeons and Candies" you are required to get descriptions of k levels from the server. Each description is a map of an n × m checkered rectangular field. Some cells of the field contain candies (each cell has at most one candy). An empty cell is denoted as "." on the map, but if a cell has a candy, it is denoted as a letter of the English alphabet. A level may contain identical candies, in this case the letters in the corresponding cells of the map will be the same.

When you transmit information via a network, you want to minimize traffic — the total size of the transferred data. The levels can be transmitted in any order. There are two ways to transmit the current level A:

1. You can transmit the whole level A. Then you need to transmit n·m bytes via the network.
2. You can transmit the difference between level A and some previously transmitted level B (if it exists); this operation requires to transmit d_A, B·w bytes, where d_A, B is the number of cells of the field that are different for A and B, and w is a constant. Note, that you should compare only the corresponding cells of levels A and B to calculate d_A, B. You cannot transform the maps of levels, i.e. rotate or shift them relatively to each other.

Your task is to find a way to transfer all the k levels and minimize the traffic.

Input

The first line contains four integers n, m, k, w (1 ≤ n, m ≤ 10; 1 ≤ k, w ≤ 1000). Then follows the description of k levels. Each level is described by n lines, each line contains m characters. Each character is either a letter of the English alphabet or a dot ("."). Please note that the case of the letters matters.

Output

In the first line print the required minimum number of transferred bytes.

Then print k pairs of integers x₁, y₁, x₂, y₂, ..., x_k, y_k, describing the way to transfer levels. Pair x_i, y_i means that level x_i needs to be transferred by way y_i. If y_i equals 0, that means that the level must be transferred using the first way, otherwise y_i must be equal to the number of a previously transferred level. It means that you will transfer the difference between levels y_i and x_i to transfer level x_i. Print the pairs in the order of transferring levels. The levels are numbered 1 through k in the order they follow in the input.

If there are multiple optimal solutions, you can print any of them.

Sample test(s)

Input

2 3 3 2
A.A
...
A.a
..C
X.Y
...

Output

14
1 0
2 1
3 1

Input

1 1 4 1
A
.
B
.

Output

3
1 0
2 0
4 2
3 0

Input

1 3 5 2
ABA
BBB
BBA
BAB
ABB

Output

11
1 0
3 1
2 3
4 2
5 1

解题思路：　以第i个level为结点i, 并增加一个0作为采用第一种方式上传level，　每个点(1...k)到标记为０的点的边权为n * m，　
　　　　　　其他从ｉ点到ｊ点的边权为diff(level i, level j)，　也就是第i个level和第ｊ个level不同的个数，
　　　　　　然后求一遍ＭＳＴ（最小生成树）即可。
Accepted Code:

 /*************************************************************************
     > File Name: 436C.cpp
     > Author: Stomach_ache
     > Mail: sudaweitong@gmail.com
     > Created Time: 2014年06月24日 星期二 17时03分36秒
     > Propose:
  ************************************************************************/

 #include <cmath>
 #include <string>
 #include <cstdio>
 #include <vector>
 #include <fstream>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 #define INF (0x3f3f3f3f)
 int n, m, k, w;
 char a[][][];
 int cost[][], d[], from[];
 bool used[];
 vector<int> OoO;

 int
 dist(int x, int y) {
       int cnt = ;
       for (int i = ; i < n; i++)
           for (int j = ; j < m; j++)
               if (a[x][i][j] != a[y][i][j]) cnt++;
     return cnt;
 }

 int
 main(void) {
       ios_base::sync_with_stdio(false);
       while (~scanf("%d %d %d %d", &n, &m, &k, &w)) {
           for (int i = ; i <= k; i++)
               for (int j = ; j < n; j++) scanf("%s", a[i][j]);

         for (int i = ; i <= k; i++) {
               for (int j = i+; j <= k; j++) {
                   cost[i][j] = cost[j][i] = dist(i, j) * w;
             }
             cost[i][] = cost[][i] = m * n;
         }
         memset(used, false, sizeof(used));
         memset(d, 0x3f, sizeof(d));
         d[] = ;
         int ans = ;
         OoO.clear();
         while (true) {
               int v = -;
             for (int u = ; u <= k; u++) {
                   if (!used[u] && (v==- || d[v] > d[u])) {
                       v = u;
                 }
             }
             if (v == -) break;
             used[v] = true;
             ans += d[v];
             for (int u = ; u <= k; u++) if (!used[u] && d[u] > cost[u][v]){
                   d[u] = cost[u][v];
                 from[u] = v;
             }
             OoO.push_back(v);
         }
         printf("%d\n", ans);
         for (int i = ; i <= k; i++) {
               printf("%d %d\n", OoO[i], from[OoO[i]]);
         }
     }

     return ;
 }