toj 4070 简单dp

相当于是在求最短路，dp之。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int INF = 1 << 29;
const int N = 21;
char road[N];
int cost[N];

void init()
{
    cost[0] = 0;
    for ( int i = 1; i < N; i++ ) cost[i] = INF;
}

bool check( int i, int j )
{
    if ( road[i] == 'R' && road[j] == 'G' ) return true;
    if ( road[i] == 'G' && road[j] == 'B' ) return true;
    if ( road[i] == 'B' && road[j] == 'R' ) return true;
    return false;
}

int main ()
{
    int t;
    scanf("%d", &t);
    while ( t-- )
    {
        scanf("%s", road);
        int len = strlen(road);
        init();
        for ( int i = 0; i < len - 1; i++ )
        {
            for ( int j = i + 1; j < len; j++ )
            {
                if ( check( i, j ) )
                {
                    int tmp = cost[i] + ( j - i ) * ( j - i );
                    if ( tmp < cost[j] ) cost[j] = tmp;
                }
            }
        }
        if ( cost[len - 1] == INF ) cost[len - 1] = -1;
        printf("%d
", cost[len - 1]);
    }
    return 0;
}