This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
using namespace std;
const int N = 405;
vector <int> pos[3];
int dp[N][N][N][4];
int cost[N][N][N][4];
int n;
int solve(int i, int j, int k, int l, int pre) {
int &res = dp[j][k][l][pre];
if (i > n)
return 0;
if (res != -1)
return res;
res = 1e9;
for (int val = 0; val < 3; val++) if (val != pre) {
int newj = j + (val == 0);
int newk = k + (val == 1);
int newl = l + (val == 2);
if (newj <= (int) pos[0].size() && newk <= (int) pos[1].size() && newl <= (int) pos[2].size()) {
res = min(res, solve(i + 1, newj, newk, newl, val) + cost[j][k][l][val]);
}
}
// cerr << i << " " << j << " " << k << " " << l << " " << pre << " " << res << "\n";
return res;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> n;
for (int i = 1; i <= n; i++) {
char c; cin >> c;
int val = 2;
if (c == 'R')
val = 0;
if (c == 'G')
val = 1;
pos[val].push_back(i);
}
for (int i = 0; i < (int) pos[0].size(); i++) {
int cur = pos[0][i];
int add = 0;
for (int j = 0; j <= (int) pos[1].size(); j++) {
add += (j > 0 && pos[1][j - 1] > cur);
int tmpadd = add;
for (int k = 0; k <= (int) pos[2].size(); k++) {
tmpadd += (k > 0 && pos[2][k - 1] > cur);
cost[i][j][k][0] = (cur + tmpadd) - (i + j + k + 1);
}
}
}
for (int j = 0; j < (int) pos[1].size(); j++) {
int cur = pos[1][j];
int add = 0;
for (int i = 0; i <= (int) pos[0].size(); i++) {
add += (i > 0 && pos[0][i - 1] > cur);
int tmpadd = add;
for (int k = 0; k <= (int) pos[2].size(); k++) {
tmpadd += (k > 0 && pos[2][k - 1] > cur);
cost[i][j][k][1] = (cur + tmpadd) - (i + j + k + 1);
}
}
}
for (int k = 0; k < (int) pos[2].size(); k++) {
int cur = pos[2][k];
int add = 0;
for (int j = 0; j <= (int) pos[1].size(); j++) {
add += (j > 0 && pos[1][j - 1] > cur);
int tmpadd = add;
for (int i = 0; i <= (int) pos[0].size(); i++) {
tmpadd += (i > 0 && pos[0][i - 1] > cur);
cost[i][j][k][2] = (cur + tmpadd) - (i + j + k + 1);
}
}
}
memset(dp, -1, sizeof(dp));
int ans = solve(1, 0, 0, 0, 3);
if (ans == 1e9)
ans = -1;
cout << ans;
return 0;
}
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