【 lightoj 1047 - Neighbor House + dp】

Description
The people of Mohammadpur have decided to paint each of their houses red, green, or blue. They’ve also decided that no two neighboring houses will be painted the same color. The neighbors of house i are houses i-1 and i+1. The first and last houses are not neighbors.

You will be given the information of houses. Each house will contain three integers “R G B” (quotes for clarity only), where R, G and B are the costs of painting the corresponding house red, green, and blue, respectively. Return the minimal total cost required to perform the work.

Input
Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case begins with a blank line and an integer n (1 ≤ n ≤ 20) denoting the number of houses. Each of the next n lines will contain 3 integers “R G B”. These integers will lie in the range [1, 1000].

Output
For each case of input you have to print the case number and the minimal cost.

Sample Input
2

4
13 23 12
77 36 64
44 89 76
31 78 45

3
26 40 83
49 60 57
13 89 99
Sample Output
Case 1: 137
Case 2: 96

有n个房子，每个房子都可以染红绿蓝三种颜色，并且给出了每个房子染每种颜色费用，相邻房子不能同色，求染完房子的最小费用。

dp[i][j]就是第i个房子染第j个颜色后的总费用基础dp～～～

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int dp[21][4],R[21],G[21],B[21];int main(){    int T,N,i,ans,nl = 0;    scanf("%d",&T);    while(T--)    {        scanf("%d",&N);        memset(dp,0,sizeof(dp));        for(i = 1 ; i <= N ; i++)        {            scanf("%d%d%d",&R[i],&G[i],&B[i]);            dp[i][0] = min(dp[i - 1][1],dp[i - 1][2]) + R[i];            dp[i][1] = min(dp[i - 1][0],dp[i - 1][2]) + G[i];            dp[i][2] = min(dp[i - 1][0],dp[i - 1][1]) + B[i];        }        ans = min(dp[N][0],min(dp[N][1],dp[N][2]));        printf("Case %d: %d\n",++nl,ans);    }    return 0;}