UVa 11552 Fewest Flops (DP)

题意：给一个字符串，把它分为ｋ块，每一块里面的字母可以任意的排序。最终字符串， 连续的一样的字母算作一个chunk，问总chunks最少是多少？

析：dp[i][j] 表示第 i 个块，第 j 位在末尾时chunk最少，状态转移方程也应该好写，如果 dp[i-1][j] 和第 i 块第一个一样，那么就总数就会-1，

否则就是直接加上。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <unordered_map>
#include <unordered_set>
#define debug() puts("++++");
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1LL << 60;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1000 + 5;
const int mod = 2000;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int dp[maxn][maxn];
char s[maxn];
int a[maxn];
bool vis[30];

int main(){
    int T;  cin >> T;
    while(T--){
        scanf("%d", &n);
        int cnt = 0;
        scanf("%s", s);
        int len = strlen(s);
        int t = len / n;
        memset(dp, INF, sizeof dp);
        memset(vis, false, sizeof vis);
        for(int i = 0; i < n; ++i)
            vis[s[i]-'a'] = true;
        for(int i = 0; i < 26; ++i)  if(vis[i])  ++cnt;
        for(int i = 0; i < n; ++i)  dp[0][i] = cnt;

        for(int i = 1; i < t; ++i){
            cnt = 0;
            memset(vis, false, sizeof vis);
            for(int j = i*n; j < (i+1)*n; ++j)
                vis[s[j]-'a'] = true;
            for(int j = 0; j < 26; ++j)  if(vis[j])  ++cnt;

            for(int j = 0; j < n; ++j){
                int fro = i * n + j;
                for(int k = 0; k < n; ++k){
                    int rear = (i-1) * n + k;
                    if(vis[s[rear]-'a'] && (1 == cnt || s[rear] != s[fro]))  dp[i][j] = min(dp[i][j], dp[i-1][k] + cnt - 1);
                    else   dp[i][j] = min(dp[i][j], dp[i-1][k] + cnt);
                }
            }
        }
        int ans = INF;
        for(int i = 0; i < n; ++i)  ans = min(ans, dp[t-1][i]);
        printf("%d
", ans);
    }
    return 0;
}