uva Fewest Flops(dp）@

Fewest Flops

Time Limit: 2000ms

Memory Limit: 131072KB

This problem will be judged on UVA. Original ID: 11552
64-bit integer IO format: %lld      Java class name: Main

Prev Submit Status Statistics Discuss Next

Font Size: + -

Type:  None Graph Theory     2-SAT     Articulation/Bridge/Biconnected Component     Cycles/Topological Sorting/Strongly Connected Component     Shortest Path         Bellman Ford         Dijkstra/Floyd Warshall     Euler Trail/Circuit     Heavy-Light Decomposition     Minimum Spanning Tree     Stable Marriage Problem     Trees     Directed Minimum Spanning Tree     Flow/Matching         Graph Matching             Bipartite Matching             Hopcroft–Karp Bipartite Matching             Weighted Bipartite Matching/Hungarian Algorithm         Flow             Max Flow/Min Cut             Min Cost Max Flow DFS-like     Backtracking with Pruning/Branch and Bound     Basic Recursion     IDA* Search     Parsing/Grammar     Breadth First Search/Depth First Search     Advanced Search Techniques         Binary Search/Bisection         Ternary Search Geometry     Basic Geometry     Computational Geometry     Convex Hull     Pick's Theorem Game Theory     Green Hackenbush/Colon Principle/Fusion Principle     Nim     Sprague-Grundy Number Matrix     Gaussian Elimination     Matrix Exponentiation Data Structures     Basic Data Structures     Binary Indexed Tree     Binary Search Tree     Hashing     Orthogonal Range Search     Range Minimum Query/Lowest Common Ancestor     Segment Tree/Interval Tree     Trie Tree     Sorting     Disjoint Set String     Aho Corasick     Knuth-Morris-Pratt     Suffix Array/Suffix Tree Math     Basic Math     Big Integer Arithmetic     Number Theory         Chinese Remainder Theorem         Extended Euclid         Inclusion/Exclusion         Modular Arithmetic     Combinatorics         Group Theory/Burnside's lemma         Counting     Probability/Expected Value Others     Tricky     Hardest     Unusual     Brute Force     Implementation     Constructive Algorithms     Two Pointer     Bitmask     Beginner     Discrete Logarithm/Shank's Baby-step Giant-step Algorithm     Greedy     Divide and Conquer Dynamic Programming                   Tag it!

[PDF Link]

Problem F

FEWEST FLOPS

A common way to uniquely encode a string is by replacing its consecutive repeating characters (or “chunks”) by the number of times the character occurs followed by the character itself. For example, the string “aabbbaabaaaa” may be encoded as “2a3b2a1b4a”. (Note for this problem even a single character “b” is replaced by “1b”.)

Suppose we have a string S and a number k such that k divides the length of S. Let S1 be the substring of S from 1 tok, S2 be the substring of S from k + 1 to 2k, and so on. We wish to rearrange the characters of each block Siindependently so that the concatenation of those permutations S’ has as few chunks of the same character as possible. Output the fewest number of chunks.

For example, let S be “uuvuwwuv” and k be 4. Then S1 is “uuvu” and has three chunks, but may be rearranged to “uuuv” which has two chunks. Similarly, S2 may be rearranged to “vuww”. Then S’, or S1S2, is “uuuvvuww” which is 4 chunks, indeed the minimum number of chunks.

Program Input

The input begins with a line containing t (1 ≤ t ≤ 100), the number of test cases. The following t lines contain an integerk and a string S made of no more than 1000 lowercase English alphabet letters. It is guaranteed that k will divide the length of S.

Program Output

For each test case, output a single line containing the minimum number of chunks after we rearrange S as described above.

INPUT

25 helloworld7 thefewestflops

OUTPUT

810

给你一个字符串，长度为k的整数倍，要你分成每个长度为k的段 每个段内可以重新编排，连续的几个字母看作一个块问最少有几个块

解：枚举每一个段中的K个字符谁当末尾，然后贪心选择，dp[i][j]表示，第i段第j个字符做末尾时的最小块

#include <iostream>#include <bits/stdc++.h>using namespace std;typedef long long LL;const int N = 1e6+7;const int inf = 0x3f3f3f3f;const LL mod  = 10000007;char str[1100];int vis[1100], dp[1007][1007];int main(){    int t;    scanf("%d", &t);    while(t--)    {        int k;        scanf("%d %s",&k, str);        int len=strlen(str);        memset(dp,0x3f3f3f3f,sizeof(dp));        for(int l=1; l<=(len/k); l++)        {            memset(vis,0,sizeof(vis));            for(int i=1; i<=k; i++)            {                int pre=(l-1)*k+i-1;                vis[str[pre]]++;            }            int cnt=0;            for(int i='a'; i<='z'; i++)                if(vis[i]) cnt++;            if(l==1)            {                for(int i=1;i<=k;i++)  dp[1][i]=cnt;                continue;            }            for(int i=1;i<=k;i++)            {                int rear=(l-1)*k+i-1;                for(int j=1;j<=k;j++)                {                    int pre=(l-2)*k+j-1;                    if(vis[str[pre]]&&(str[pre]!=str[rear]||cnt==1)) dp[l][i]=min(dp[l][i],dp[l-1][j]+cnt-1);                    else dp[l][i]=min(dp[l][i],dp[l-1][j]+cnt);                }            }        }        int ans=0x3f3f3f3f;        for(int i=1;i<=k;i++) ans=min(ans,dp[len/k][i]);        cout<<ans<<endl;    }    return 0;}