tju 4070 简单dp

4070.   Road

---

Time Limit: 1.0 Seconds   Memory Limit:65536K
Total Runs: 84   Accepted Runs:76

---

There is a one-dimensional road. The road is separated into N consecutive parts. The parts are numbered 0 through N-1, in order. MJ is going to walk from part 0 to part N-1.

MJ also noticed that each part of the road has a color: either red, green, or blue. Part 0 is red.

MJ is going to perform a sequence of steps. Each step must lead in the positive direction. That is, if his current part is i, the next step will take him to one of the parts i+1 through N-1, inclusive. His steps can be arbitrarily long. However, longer steps are harder: a step of length j costs j*j energy.

Additionally, MJ wants to step on colors in a specific order: red, green, blue, red, green, blue, ... That is, he starts on the red part 0, makes a step to a green part, from there to a blue part, and so on, always repeating red, green, and blue in a cycle. Note that the final part N-1 also has some color and thus MJ must reach it in a corresponding step.

You are given a String road containing N elements. For each i, element i of road is the color of part i: 'R' represents red, 'G' green, and 'B' blue. If MJ can reach part N-1 in the way described above, return the smallest possible total cost of doing so. Otherwise, return -1.

Input

First line is a integer T, the number of test cases.

The next T lines, each line is a string S represent the road.The length of S is no longer than 20.

Output

For each test case, output the smallest possible total cost or -1 if it's impossible.

Sample Input

4RGGGBRGBRGBRGBRBBGGGRRRBRRBGGGBBBBR

Sample Output

88-150

Source: TJU Team Selection 2014 Round1

题意：每次走步的花费为长度的平方，必须按照红绿蓝的顺序走，问走到路的尽头的最小花费

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#define MAX 27using namespace std;int n;char road[MAX];int dp[MAX][MAX];int get ( char ch ){    if ( ch == 'R' ) return 0;    if ( ch == 'G' ) return 1;    else return 2;}int main ( ){    scanf ( "%d" , &n );    while ( n -- )    {        scanf ( "%s" , road );        memset ( dp , 0x3f , sizeof ( dp ) );        int INF = dp[0][0];        int ans = INF;        dp[0][0] = 0;        int len = strlen ( road );        for ( int i = 1 ; i <= len ; i++ )            for ( int j = i ; j < len ; j++ )            {                if ( get ( road[j] ) == i%3 )                    for ( int k = 0 ; k < j ; k++ )                        if ( get ( road[k] ) == (i-1)%3 )                            dp[i][j] = min ( dp[i][j] , dp[i-1][k] + (j-k)*(j-k) );                if ( j == len-1 ) ans = min ( ans , dp[i][j] );            }        if ( ans >= INF ) puts ( "-1" );        else printf ( "%d\n" , ans );       }}