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:

 1 /*************************************************************************  2  > File Name: 436C.cpp  3  > Author: Stomach_ache  4  > Mail: [email protected]  5  > Created Time: 2014年06月24日 星期二 17时03分36秒  6  > Propose:  7  ************************************************************************/

 8 

 9 #include <cmath>

10 #include <string>

11 #include <cstdio>

12 #include <vector>

13 #include <fstream>

14 #include <cstring>

15 #include <iostream>

16 #include <algorithm>

17 using namespace std; 18 

19 #define INF (0x3f3f3f3f)

20 int n, m, k, w; 21 char a[1002][12][12]; 22 int cost[1002][1002], d[1002], from[1002]; 23 bool used[1002]; 24 vector<int> OoO; 25 

26 int

27 dist(int x, int y) { 28       int cnt = 0; 29       for (int i = 0; i < n; i++) 30           for (int j = 0; j < m; j++) 31               if (a[x][i][j] != a[y][i][j]) cnt++; 32     return cnt; 33 } 34 

35 int

36 main(void) { 37       ios_base::sync_with_stdio(false); 38       while (~scanf("%d %d %d %d", &n, &m, &k, &w)) { 39           for (int i = 1; i <= k; i++) 40               for (int j = 0; j < n; j++) scanf("%s", a[i][j]); 41         

42         for (int i = 1; i <= k; i++) { 43               for (int j = i+1; j <= k; j++) { 44                   cost[i][j] = cost[j][i] = dist(i, j) * w; 45  } 46             cost[i][0] = cost[0][i] = m * n; 47  } 48         memset(used, false, sizeof(used)); 49         memset(d, 0x3f, sizeof(d)); 50         d[0] = 0; 51         int ans = 0; 52  OoO.clear(); 53         while (true) { 54               int v = -1; 55             for (int u = 0; u <= k; u++) { 56                   if (!used[u] && (v==-1 || d[v] > d[u])) { 57                       v = u; 58  } 59  } 60             if (v == -1) break; 61             used[v] = true; 62             ans += d[v]; 63             for (int u = 0; u <= k; u++) if (!used[u] && d[u] > cost[u][v]){ 64                   d[u] = cost[u][v]; 65                 from[u] = v; 66  } 67  OoO.push_back(v); 68  } 69         printf("%d\n", ans); 70         for (int i = 1; i <= k; i++) { 71               printf("%d %d\n", OoO[i], from[OoO[i]]); 72  } 73  } 74 

75     return 0; 76 }