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:

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 #define MS(a) memset(a,0,sizeof(a))
 5 #define MP make_pair
 6 #define PB push_back
 7 const int INF = 0x3f3f3f3f;
 8 const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
 9 inline ll read(){
10     ll x=0,f=1;char ch=getchar();
11     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
12     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
13     return x*f;
14 }
15 //////////////////////////////////////////////////////////////////////////
16 const int maxn = 1e5+10;
17 
18 ll cost[30][30];
19 
20 int main(){
21     for(int i=0; i<26; i++){
22         for(int j=0; j<26; j++){
23             cin >> cost[i][j];
24         }
25         cost[i][i] = 0;
26     }
27     for(int k=0; k<26; k++)
28         for(int i=0; i<26; i++){
29             for(int j=0; j<26; j++)
30                 cost[i][j] = min(cost[i][j],cost[i][k]+cost[k][j]);
31         }
32 
33     string s;
34     cin >> s;
35     ll len = s.size();
36     ll ans = 0;
37 
38     // int p,pp,ppp;
39     for(int i=0; i<len/2; i++){
40         ll minn = INF;
41         int s1 = s[i]-'a';
42         int s2 = s[len-i-1]-'a';
43         // cout << s1 << " " << s2 << endl;
44         for(int j=0; j<26; j++){
45             if(minn>cost[s1][j]+cost[s2][j]){
46                 minn = cost[s1][j]+cost[s2][j];
47                 // p=s1, pp=j, ppp=s2;
48             }
49         }
50         ans += minn;
51     }
52 
53     // cout << p << " " << pp << "   " << ppp << " " << pp << endl;
54 
55     cout << ans << endl;
56 
57     return 0;
58 }
59 
60 // http://codeforces.com/gym/100733/problem/H