Codeforces Gym 100733H Designation in the Mafia flyod

Designation in the Mafia
Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=88994#problem/H

Description

The Shitalian mafia has a very peculiar way to name a new associate. The newbie goes through tests to measure his strength, agility, sagacity and influence. The size of the nickname is defined by the results of the tests.

Then, the newbie chooses the characters of his nickname, that is, letters from the alphabet. But Shi has nothing better to do, so he demands that the nickname is transformed in a palindrome.

A palindrome is a string that can be read the same way if we reverse it. For instance BANANAB is a palindrome, and BANANAS is not.

Everything in Shitalia has a cost, including transforming characters. You will receive a matrix P, of size 26x26. The element P_ij is the cost of transforming the i-th letter of the alphabet into the j-th letter of the alphabet. You can apply as many transformations as you want.

Find the minimum cost to transform the nickname into a palindrome.

Input

The first 26 lines form the matrix P. Each line i contains exactly 26 integers P_ij(0 ≤ P_ij ≤ 10⁶), indicating . The last line contains the nickname that the newbie has chosen, which is a string with n(1 ≤ n ≤ 10⁶) lowercase letters.

Output

Print the minimal cost to transform the nickname into a palindrome.

Sample Input

9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
xx

Sample Output

0

HINT

题意

给你每一个字符变成另外一个字符的花费

然后问你最小需要多少才能把这个字符串变成回文串

题解：

看懂第三个样例，基本这道题就出来了

注意要跑flyod，这个cost[x][y]不一定比cost[x][k]+cost[k][y]低

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200051
#define mod 10007
#define eps 1e-9
int Num;
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//**************************************************************************************

long long cost[30][30];
int main()
{
    for(int i=0;i<26;i++)
    {
        for(int j=0;j<26;j++)
        {
            cin>>cost[i][j];
        }
        cost[i][i]=0;
    }
    for(int k=0;k<26;k++)
    {
        for(int i=0;i<26;i++)
        {
            for(int j=0;j<26;j++)
            {
                cost[i][j]=min(cost[i][j],cost[i][k]+cost[k][j]);
            }
        }
    }
    string s;
    cin>>s;
    ll ans=0;
    int len = s.size();

    for(int i=0;i<len/2;i++)
    {
        ll mins = inf;
        int i1 = s[i]-'a';
        int i2 = s[len-i-1] - 'a';
        for(int j=0;j<26;j++)
            mins = min(mins,cost[i1][j]+cost[i2][j]);
        ans+=mins;
    }
    cout<<ans<<endl;
}