Codeforces Gym 100733H Designation in the Mafia flyod

题目链接:

http://codeforces.com/gym/100733/problem/H

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:

懂第三个样例，基本这道题就出来了(a->b->c)
注意要跑flyod，这个cost[x][y]不一定比cost[x][k]+cost[k][y]低

代码:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MS(a) memset(a,0,sizeof(a))
#define MP make_pair
#define PB push_back
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
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;
}
//////////////////////////////////////////////////////////////////////////
const int maxn = 1e5+10;

ll 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 len = s.size();
    ll ans = 0;

    // int p,pp,ppp;
    for(int i=0; i<len/2; i++){
        ll minn = INF;
        int s1 = s[i]-'a';
        int s2 = s[len-i-1]-'a';
        // cout << s1 << " " << s2 << endl;
        for(int j=0; j<26; j++){
            if(minn>cost[s1][j]+cost[s2][j]){
                minn = cost[s1][j]+cost[s2][j];
                // p=s1, pp=j, ppp=s2;
            }
        }
        ans += minn;
    }

    // cout << p << " " << pp << "   " << ppp << " " << pp << endl;

    cout << ans << endl;

    return 0;
}

// http://codeforces.com/gym/100733/problem/H