- 回文(palindrome)
40分
- @ 2026-4-15 16:58:56
using namespace std;
int main(){
int n;
cin>>n;
vector<int> v(26);
for(int i=0;i<26;++i)cin>>v[i];
string s;
cin>>s;
vector<vector<int>> m(26, vector<int>(26));
for (int x=0;x<26;++x)
for (int y=0;y<26;++y) {
int k=INT_MAX;
for(int c=0;c<26;++c){
int t=0;
if(x!=c)t+=v[c];
if(y!=c)t+=v[c];
k=min(k, t);
}
m[x][y]=k;
}
int h=n/2;
vector<pair<int,int>> p;
vector<int>o;
int b=0;
for(int i=0;i<h;++i){
int j=n-1-i;
int a=s[i]-'a';
int d=s[j]-'a';
p.push_back(make_pair(a, d));
int c=m[a][d];
o.push_back(c);
b+=c;
}
int g=0;
for(int i=0;i<h;++i){
int a=p[i].first;
int d=p[i].second;
int o_i=o[i];
int n_c=m[d][a];
g=max(g, o_i - n_c);
}
for(int i=0;i<h;++i)
for (int j=i+1;j<h;++j) {
int a=p[i].first;
int d=p[i].second;
int b_p=p[j].first;
int c_p=p[j].second;
int o_i=o[i], o_j=o[j];
int o_t=o_i+o_j;
int c1=m[b_p][d]+m[a][c_p];
int g1=o_t-c1;
int c2=m[c_p][d]+m[a][b_p];
int g2=o_t-c2;
int c3=m[b_p][a]+m[d][c_p];
int g3=o_t-c3;
int c4=m[c_p][a]+m[d][b_p];
int g4=o_t-c4;
g=max(g,max(max(g1,g2),max(g3,g4)));
}
cout<<max(0,b-g)<<endl;
return 0;
}
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Information
- ID
- 1611
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 9
- Tags
- (None)
- # Submissions
- 176
- Accepted
- 12
- Uploaded By