CodeForces 73C LionAge II

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本文介绍了一道关于字符串优化的问题，通过动态规划算法解决如何在限定更改次数内最大化字符串评分。文章详细阐述了解题思路及代码实现。

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Vasya plays the LionAge II. He was bored of playing with a stupid computer, so he installed this popular MMORPG, to fight with his friends. Vasya came up with the name of his character — non-empty string s, consisting of a lowercase Latin letters. However, in order not to put up a front of friends, Vasya has decided to change no more than k letters of the character name so that the new name sounded as good as possible. Euphony of the line is defined as follows: for each pair of adjacent letters x and y (x immediately precedes y) the bonus c(x, y) is added to the result. Your task is to determine what the greatest Euphony can be obtained by changing at most k letters in the name of the Vasya's character.

Input

The first line contains character's name s and an integer number k (0 ≤ k ≤ 100). The length of the nonempty string s does not exceed 100. The second line contains an integer number n (0 ≤ n ≤ 676) — amount of pairs of letters, giving bonus to the euphony. The next n lines contain description of these pairs «x y c», which means that sequence xy gives bonus c (x, y— lowercase Latin letters, - 1000 ≤ c ≤ 1000). It is guaranteed that no pair x y mentioned twice in the input data.

Output

Output the only number — maximum possible euphony оf the new character's name.

Sample test(s)

input
winner 4
4
s e 7
o s 8
l o 13
o o 8

output
36

input
abcdef 1
5
a b -10
b c 5
c d 5
d e 5
e f 5

output
20

Note

In the first example the most euphony name will be looser. It is easy to calculate that its euphony is 36.

题意：给你一个串，你一次操作可以改变任意一个字符，但是你的操作次数最多为k次。再给出n对

代表这两个字符如果相邻，那么会产生c的价值。问最后你可以得到的最多价值是多少。

思路：很明显是一道dp的题目，于是就开始想dp的状态，首先给定字符串的长度一定是要的，操作

次数也是要的，但是这样还是不够解。于是就想了一个三维的dp；

dp[pos][k][i]: pos--代表第pos位，k--代表改变了k次，i代表把pos这个位子的字符改成i；

状态转移从当前pos位置状态推出pos+1位置的状态。这时候只需要再for一个下一个状态的字符是什

么，就可以进行转移了，如果下一个状态的字符就是原串的字符那么k不改变，如果不是那么k+1;

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<iostream>
#include<stdlib.h>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<bitset>
#pragma comment(linker, "/STACK:1024000000,1024000000")
template <class T>
bool scanff(T &ret){
    char c; int sgn; T bit=0.1;
    if(c=getchar(),c==EOF) return 0;
    while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
    if(c==' '||c=='\n'){ ret*=sgn; return 1; }
    while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;
    ret*=sgn;
    return 1;
}
#define inf 1073741824
#define llinf 4611686018285162540LL
#define eps 1e-8
#define mod 9223372034707292160LL
#define pi acos(-1.0)
#define lth (th<<1)
#define rth (th<<1|1)
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define drep(i,a,b) for(int i=a;i>=b;i--)
#define mset(x,val) memset(x,val,sizeof(x))
#define mcpy(x,y) memcpy(x,y,sizeof(y))
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define fuck puts("FUCK!")
#define mpii(a,b) make_pair(a,b);
#define NN 101010
#define MM 202020
using namespace std;
typedef unsigned long long ll;
typedef long double lb;
typedef pair<int,int> pii;
int dp[111][111][111],kk,len,n,val;
char s[111],c,d;
int a[268][268];
int main(){
    scanf("%s%d",s,&kk);
    scanf("%d%*C",&n);
    len=strlen(s);
    mset(dp,0);
    mset(a,0);
    rep(i,1,n){
        scanf("%c%*C%c",&c,&d);
        scanf("%d%*C",&val);
        a[c][d]=val;
    }

    rep(pos,0,len){
        rep(k,0,kk){
            for(int i='a';i<='z';i++){
                dp[pos][k][i]=-inf;
            }
        }
    }

    for(int i='a';i<='z';i++) dp[0][1][i]=0;
    dp[0][0][s[0]]=0;

    rep(pos,0,len){
        rep(k,0,kk){
            for(int i='a';i<='z';i++){
                if(dp[pos][k][i]==-inf) continue;
                for(int j='a';j<='z';j++){
                    int tmp=1;
                    if(s[pos+1]==j)tmp=0;
                    dp[pos+1][k+tmp][j]=max(dp[pos+1][k+tmp][j],dp[pos][k][i]+a[i][j]);
                }
            }
        }
    }

    int ans=-inf;
    for(int i=0;i<=kk;i++){
        for(int j='a';j<='z';j++){
            ans=max(ans,dp[len-1][i][j]);
        }
    }
    printf("%d\n",ans);
    return 0;
}