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Copy pathMinString(DP).cpp
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133 lines (111 loc) · 4.39 KB
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Copy pathMinString(DP).cpp
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133 lines (111 loc) · 4.39 KB
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#include "algorithm"
#include "iostream"
#include "numeric"
#include "iomanip"
#include "cstring"
#include "math.h"
#include "bitset"
#include "string"
#include "vector"
#include "ctime"
#include "queue"
#include "stack"
#include "map"
#include "set"
#include "ext/pb_ds/assoc_container.hpp" // Common file
#include "ext/pb_ds/tree_policy.hpp" // Including tree_order_statistics_node_update
#include "ext/pb_ds/detail/standard_policies.hpp"
using namespace std;
using namespace __gnu_pbds;
#define f first
#define lgn 25
#define endl '\n'
#define sc second
#define pb push_back
#define N (int) 200+5
#define PI acos(-1.0)
#define int long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define eb emplace_back
#define mii map<int,int>
#define vpii vector<pii>
#define pii pair<int,int>
#define pq priority_queue
#define BLOCK (int)sqrt(N)
#define test(x) while(x--)
#define all(x) begin(x),end(x)
#define allr(x) x.rbegin(),x.rend()
#define fo(i,a,n) for(int i=a;i<n;i++)
#define rfo(i,n,a) for(int i=n;i>=a;i--)
#define FAST ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define time() cerr << "Time : " << (double)clock() / (double)CLOCKS_PER_SEC << "s\n"
#define bug(...) __f (#__VA_ARGS__, __VA_ARGS__)
typedef tree< int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update >
OS ;
template <typename Arg1>
void __f (const char* name, Arg1&& arg1) { cout << name << " : " << arg1 << endl; }
template <typename Arg1, typename... Args>
void __f (const char* names, Arg1&& arg1, Args&&... args)
{
const char* comma = strchr (names + 1, ',');
cout.write (names, comma - names) << " : " << arg1 << " | "; __f (comma + 1, args...);
}
const int inf = 0x3f3f3f3f;
const int INF = 0x3f3f3f3f3f3f3f3f;
int n,m,k,q;
string s;
int dp[N][N]; // dp[i][j] -> denotes minimum length we can construct by changing in interval (i,j)
int val[N][N]; // val[i][j] -> It is used since if we get (a,b) then we can change it to c
// and let suppose a -> 1st index and b-> 2nd index , so now val[1][2] will be c
void go()
{
cin >> s;
n = s.size();
s = " " + s;
memset( dp , 0, sizeof dp );
memset( val , 0 , sizeof val );
for( int i = 1; i <= n; i++)
{
for( int j = i; j <= n; j++)
{
if( i == j )
{
if( s[i] == 'a' ) val[i][i] = 1;
if( s[i] == 'b' ) val[i][i] = 2;
if( s[i] == 'c' ) val[i][i] = 3;
}
dp[i][j] = ( j - i + 1);
}
}
for(int gap = 1; gap <= n; gap++)
{
for( int i = 1; i + gap <= n; i++)
{
int j = i + gap;
for( int k = i; k < j; k++)
{
dp[i][j] = min( dp[i][j] , dp[i][k] + dp[k+1][j] );
int sum = val[i][k] + val[k+1][j];
if( val[i][k] and val[k+1][j] and val[i][k] != val[k+1][j] and dp[i][j] > 0) // if values are different then we can reduce the length
{
if( sum == 3 ) val[i][j] = 3; // here we get ( 1,2) or (2 ,1), so we change it to -> 3
if( sum == 4 ) val[i][j] = 2; // here we get ( 1,3) or (3 ,1), so we change it to -> 2
if( sum == 5 ) val[i][j] = 1; // here we get ( 3,2) or (2 ,3), so we change it to -> 1
dp[i][j] = min( dp[i][j] , max(1LL , dp[i][j] - 1) ); // either the answer is (i,k) + (k+1,j)
// which we have already calculated before finding sum
// so now the length will reduce by 1 , so to avoid zero length max is used
}
}
}
}
cout << dp[1][n] << endl;
}
int32_t main()
{
FAST;
int t=1;
cin>>t;
test(t) go();
}