LightOJ 1047 Neighbor House

dp[i][j]：涂前i的房子， 第i间房涂第j种颜色的最小花费

#include <cstdio>#include <algorithm>using namespace std;int main(){    int T;    scanf("%d", &T);    for(int ca = 1; ca <= T; ca++)    {        int n, dp[25][3];        scanf("%d", &n);        for(int i = 1; i <= n; i++)            scanf("%d%d%d", &dp[i][0], &dp[i][1], &dp[i][2]);        for(int i = 2; i <= n; i++)        {            dp[i][0] += min(dp[i - 1][1], dp[i - 1][2]);            dp[i][1] += min(dp[i - 1][0], dp[i - 1][2]);            dp[i][2] += min(dp[i - 1][0], dp[i - 1][1]);        }        printf("Case %d: %d\n", ca, min(min(dp[n][0], dp[n][1]), dp[n][2]));    }    return 0;}