Sicily 1627. Text Messaging Improvement?

1627. Text Messaging Improvement?

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB , Special Judge

Description

On a standard mobile phone the letters are distributed across the keys 2 through 9 as:

23ABCDEF456GHIJKLMNO789PQRSTUVWXYZ

To enter the letter C, you press key 2 three times (seeing A-B-C). The number of keystrokes to enter a letter depends on where it is in the list of letters on its key.

The Flathead Telephone Company (FTC) is considering rearranging the letters on the keys to reduce the average number of keystrokes required to enter names etc. or send text messages. The letters must still appear in alphabetical order on the keys but different numbers of letters may appear on each key and possibly more keys could be used. FTC has several databases of letter frequencies used in different applications. For instance, it might help to move S from the 7 key to the 8 key. They need a program which is given the frequencies of the letters and a number of keys and returns the assignment of letters to keys with the smallest average number of keystrokes using the given frequencies. Each key used must have at least one letter and at most eight letters.

Input

The first line of input contains a single integer N , (1≤N≤1000) which is the number of data sets that follow. Each data set consists of three lines of input. The first line contains a single integer K , (4≤K≤26) , the number of keys which are to be used. The second and third lines contain 13 decimal values each giving the percent frequency of the letters A through Z in order.

Output

For each data set, you should generate one line of output with the following values: The data set number as a decimal integer (start counting at one), the best average number of keystrokes to three decimal places, a space and the letters A through Z, for the best arrangement, in order with a single space at the break between letters on different keys. It is possible that the same input data set may produce different output.

Sample Input

2 8 8.167 1.492 2.782 4.253 12.702 2.228 2.015 6.094 6.966 0.153 0.772 4.025 2.4066.749 7.507 1.929 0.095 5.987 6.327 9.056 2.758 0.978 2.360 0.150 1.974 0.075 9 1.0 10.0 11.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 10.0 10.0 10.0 10.0 10.0 11.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0

Sample Output

1 1.647 AB CD EFG HIJK LM NOPQ RS TUVWXYZ 
2 1.570 A B CDEFG HIJKLM N OP QR STUV WXYZ
// Problem#: 1627// Submission#: 3593577// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University/* * Text messaging solution * by Fred Pickel * October 2008 */#include <stdio.h>#include <stdlib.h>#include <math.h>#define MAX_PROBS   1000#define EPS .001#define ERR .001char inbuf[512];int nButtons;double charFreq[26];double freqDenom;int keyLetterCounts[26];int ReadDataSet(int index){    int i;    if(fgets(&(inbuf[0]), 255, stdin) == NULL)    {        fprintf(stderr, "read failed on problem %d num buttons\n", index);        return -1;    }    inbuf[511] = 0;    if(sscanf(inbuf, "%d", &nButtons) != 1)    {        fprintf(stderr, "scan failed on problem %d num buttons\n", index);        return -2;    }    if(fgets(&(inbuf[0]), 255, stdin) == NULL)    {        fprintf(stderr, "read failed on problem %d char freq a-m\n", index);        return -3;    }    inbuf[511] = 0;    if(sscanf(inbuf, "%lf %lf %lf %lf %lf %lf %lf %lf %lf %lf %lf %lf %lf ",         &charFreq[0], &charFreq[1], &charFreq[2], &charFreq[3], &charFreq[4],        &charFreq[5], &charFreq[6], &charFreq[7], &charFreq[8], &charFreq[9],        &charFreq[10], &charFreq[11], &charFreq[12]) != 13)    {        fprintf(stderr, "scan failed on problem %d  char freq a-m\n", index);        return -4;    }    if(fgets(&(inbuf[0]), 255, stdin) == NULL)    {        fprintf(stderr, "read failed on problem %d char freq n-z\n", index);        return -5;    }    inbuf[511] = 0;    if(sscanf(inbuf, "%lf %lf %lf %lf %lf %lf %lf %lf %lf %lf %lf %lf %lf ",         &charFreq[13], &charFreq[14], &charFreq[15], &charFreq[16], &charFreq[17],        &charFreq[18], &charFreq[19], &charFreq[20], &charFreq[21], &charFreq[22],        &charFreq[23], &charFreq[24], &charFreq[25]) != 13)    {        fprintf(stderr, "scan failed on problem %d  char freq n-z\n", index);        return -6;    }    for(i = 0, freqDenom = 0.0; i < 26; i++)    {        freqDenom += charFreq[i];    }    return 0;}// bestAvg[i][j] = best avg num key presses using i+1 keys to do letters 0 to jdouble bestAvg[26][26];// bestLetterCount = num letters on key i+1 to get value in bestAvgint bestLetterCount[26][26];    /* solving: * we start with the first key used and count keystrokes using 1-8 letters * then we look at key 2, for each letter count on key 1 we add more  letters on key 2 * choosing the smallest total count etc *  */int TeleSolve(){    int i, j, k, lim;    double curCount;    // init    for(i = 0; i < 26; i++)    {        for(j = 0; j < 26; j++)        {            bestAvg[i][j] = 5000.0; // bigger than any valid count            bestLetterCount[i][j] = 0;        }    }    // do first key    for(i = 0, curCount = 0.0; i < 8 ; i++) // max 8 on a key    {        curCount += (i+1)*charFreq[i];  // cost of another letter on key        bestAvg[0][i] = curCount;        bestLetterCount[0][i] = i+1;    }    // now do remaining keys    for(j = 1; j < nButtons ; j++)    {        for(k = j; k < 26; k++) // at least one letter on each prior key        {   // for each combination of preiously compute sums            // start from there with up to 8 keys, choosing best arrangement            if(bestAvg[j-1][k-1] > 4000.0)            {   // cannot be useful, would be more that 8 chars on prior keys                break;            }            lim = 26 - nButtons + j + 1;    // cannot go beyond this, leave room for one keys on each later button            if(lim > 8) lim = 8;    // max 8 on a key            for(i = 0, curCount = bestAvg[j-1][k-1]; i < lim ; i++)            {   // for each possible letter on this key startingwith letter 'A'+k                // see what cost of i letters on this key would add, if better than prior, remeber                curCount += (i+1)*charFreq[i+k];                if(bestAvg[j][k+i] > curCount)                {                    bestAvg[j][k+i] = curCount;                    bestLetterCount[j][k+i] = i+1;                }            }        }    }    return 0;}void PrintSolution(int probNum){    int i, j, k;    // scan backwards through the bestLetterCounts array to find out    // how many letters on each key    for(i = nButtons - 1, j = 25; i >= 0 ; i--)    {        keyLetterCounts[i] = bestLetterCount[i][j];        j -= keyLetterCounts[i];    }    // print prob num and best avg key presses    printf("%d %.3f ", probNum, bestAvg[nButtons-1][25]/100.0);    // print letters on keys    for(i = 0, j=0; i < nButtons ; i++)    {        for(k = 0; k < keyLetterCounts[i] ; k++)        {            putchar('A' + j);            j++;        }        putchar(' ');    }    printf("\n");}        int main(){    int probCnt, curProb, ret;    if(fgets(&(inbuf[0]), 255, stdin) == NULL)    {        fprintf(stderr, "read failed on problem count\n");        return -1;    }    inbuf[255] = 0;    probCnt = atoi(&(inbuf[0]));    if((probCnt < 1) || (probCnt > MAX_PROBS))    {        fprintf(stderr, "Problem count %d not in range 1...%d\n", probCnt, MAX_PROBS);        return -2;    }    for(curProb = 1; curProb <= probCnt ; curProb++)    {        if((ret = ReadDataSet(curProb)) != 0)        {            fprintf(stderr, "ReadDataSet returned %d on problem %d\n", ret, curProb);            return -3;        }        TeleSolve();        PrintSolution(curProb);    }    return 0;}