2014-06-03

UVa 12721 - Cheap B-Subsequence

1. Problem
2. Input
3. Output
4. Sample Input
5. Sample Output
6. Solution

Problem

Some time ago, Dejan Stojanovic, a Serbian poet, said: “Words rich in meaning can be cheap in sound
effects.” Is it true? A String Processing professor at UFPE wants to test this quote with strings. For that, he defined what he calls a “cheap B-subsequence”. A cheap B-subsequence, according to his definition, is a subsequence of size B, of a string S(B <= |S|), that has the lowest associated cost. To define the cost of a string, the professor determined a series of rules to each letter of the alphabet. The alphabet that he used contains only lowercase letters. The rule of a letter is defined as a set of pairs (P_i, C_i), which indicates that if this letter appears in a position X on the subsequence, where X is a multiple of P_i, then the cost of (X/P_i) * C_i will be added to the total cost of this subsequence. Let’s show an example. Suppose we have the following rules:

[a] = {(2, 3), (4, 10)}
[b] = {(1, 4), (7, 50)}
[c] = {(1, 2), (4, 20)}
[d..z] = { } // there are no rules for the characters `d` to `z`

Suppose we have the string abaabcbc, and B = 4. If we choose the subsequence aabc (abaabcbc), we would do the following procedure to calculate the associated cost:

The first letter of the sequence is an a, and the position 1 is neither multiple of 2 or 4, so the cost is 0;
The second letter of the sequence is another a, and the position 2 is a multiple of 2, so we’ll add the cost of (2/2) * 3 = 3;
The third letter of the sequence is a b, and the position 3 is multiple of 1, so we will add the cost of (3/1) * 4 = 12;
The last letter of the sequence is a c, and the position 4 is a multiple of 1 and 4, so we will add
the cost of (4/1) * 2 + (4/4)* 20 = 28.

The total associated cost to this subsequence is 43, which is not the lowest cost, since we could have chosen aaab (abaabcbc) and obtained an associated cost of 19 | this is indeed the cost of the cheap B-subsequence. Given the string S and the integer B, and the rules of the alphabet, your task is to create a program that tells the professor the cost of the cheap B-subsequence.

Input

The first line contains T (T < 100)－　the number of test cases, after this line T test cases follows. The first line of a test case contains a string S of lowercase letters and an integer B
(1 <= B <= |S| <= 100). Each of the next 26 lines describe the rule of each letter. The first of the 26 lines corresponds to the rule of the letter a; the following line corresponds to the rule of the letter b; the last of the 26 lines corresponds to the rule of the letter z. Each line containing a rule is described in the following way:

Q, P1, C1, P2, C2, … PQ, CQ

(1 <= Q <= 10; 1 <= Pi <= |S|; 1 <= Ci <= 50), where Q is the amount of pairs associated to this rule, and is followed by the pairs themselves.

Output

For each test case print a line containing Case #X: Y, where X is the case number, starting at 1, and Y is the cost of the cheap B-subsequence.

Sample Input

2
abcd 1
1 1 20
1 1 15
1 1 8
1 1 30
1 1 2
0 (21 lines)
abaabcbc 4
2 2 3 4 10
2 1 4 7 50
2 1 2 4 20
0 (23 lines)

Sample Output

1 2	Case #1: 8 Case #2: 19

Solution

題目描述：

找一個長度為 B 的 subsequence，並且每一個字符成本如題目所定義，求最小構成成本。

題目解法：

動態規劃，狀態 dp[length][subsequence_index]

#include <stdio.h>
#include <string.h>
#include <vector>
#include <algorithm>
using namespace std;
int dp[128][128], cost[26][128];
int main() {
    int testcase, cases = 0;
    char S[128];
    int B, P, C, N, M;
    scanf("%d", &testcase);
    while(testcase--) {		
        scanf("%s %d", S+1, &B);
        
        N = strlen(S+1);
        memset(dp, 0x3f, sizeof(dp));
        memset(cost, 0, sizeof(cost));
        
        for(int i = 0; i < 26; i++) {
            scanf("%d", &M);
            for(int j = 0; j < M; j++) {
                scanf("%d %d", &P, &C);
                int oP = P;
                while(oP <= N)
                    cost[i][oP] += oP / P * C, oP += P;
            }
        }		
        dp[0][0] = 0;
        for(int i = 0; i < B; i++) {
            for(int j = 0; j <= N; j++) {
                for(int k = j + 1; k <= N; k++) {
                    int c = cost[S[k] - 'a'][i + 1];
                    dp[i+1][k] = min(dp[i+1][k], dp[i][j] + c);
                }
            }
        }
        int ret = 0x3f3f3f3f;
        for(int i = 1; i <= N; i++)
            ret = min(ret, dp[B][i]);
        printf("Case #%d: %d\n", ++cases, ret);
    }
    return 0;
}
/*
2
abcd 1
1 1 20
1 1 15
1 1 8
1 1 30
1 1 2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
abaabcbc 4
2 2 3 4 10
2 1 4 7 50
2 1 2 4 20
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
*/

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