Codeforces 436 C. Dungeons and Candies

MST ....

C. Dungeons and Candiestime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

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

When you transmit infZ records? Http://www.bkjia.com/kf/ware/vc/ "target =" _ blank "class =" keylink "> latency + QTwvZW0 + latency + q1_vzw0 + latency + ZDwvZW0 + latency/latency + QTwvZW0 + latency + q1_vzw0 + latency + QTwvZW0 + LD88ZW0 + pushed/pushed + h3D88ZW0 + bjwvZW0 + LD88ZW0 + pushed/odw/pushed + bjwvZW0 + pushed = ". "). please note that the case of the letters matters.

Output

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

Then printKPairs of integersX1 ,?Y1 ,?X2 ,?Y2 ,?...,?XK,?YK, Describing the way to transfer levels. PairXI,YIMeans that levelXINeeds to be transferred by wayYI. IfYIEquals 0, that means that the level must be transferred using the first way, otherwiseYIMust be equal to the number of a previusly transferred level. It means that you will transfer the difference between levelsYIAndXITo transfer levelXI. Print the pairs in the order of transferring levels. The levels are numbered 1 throughKIn 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 2A.A...A.a..CX.Y...

Output

141 02 13 1

Input

1 1 4 1A.B.

Output

31 02 04 23 0

Input

1 3 5 2ABABBBBBABABABB

Output

111 03 12 34 25 1

#include 
 
  #include 
  
   #include 
   
    #include #include 
    
     using namespace std;struct Edge{    int u,v,w;}edge[1100*1100];bool cmp(Edge a,Edge b){    return a.w
     
       vc[1100];int get_diff(char a[][20],char b[][20] ){    int ret=0;    for(int i=0;i