题目链接：?problem=1047 题意：求(p[i][j])上下相邻的 j 不能相同的数塔的最小和。 解法：看代码！ 代码：
;int t;int n;int p[25][5];int dp[25][25];int pos[25];int main(){cin >> t;for (int ca = 1; ca <= t; ca++){memset(dp,0,sizeof(dp));cin >> n;for (int i = 1; i <= n; i++)for (int j = 1; j <= 3; j++)cin >> p[i][j];for (int i = 1; i <= n; i++)for (int j = 1; j <= 3; j++){if (j == 1)dp[i][j] = min(dp[i – 1][2], dp[i – 1][3]) + p[i][j];if (j == 2)dp[i][j] = min(dp[i – 1][1], dp[i – 1][3]) + p[i][j];if (j == 3)dp[i][j] = min(dp[i – 1][2], dp[i – 1][1]) + p[i][j];}cout << “Case ” << ca << “: ” << min(dp[n][1] ,min(dp[n][2],dp[n][3]))<< endl;}return 0;}
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