LightOJ 1047 Neighbor House (线性dp 类数字三角形)

1047 - Neighbor House

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Time Limit: 0.5 second(s)Memory Limit: 32 MB

The people of Mohammadpur have decided to paint each oftheir houses red, green, or blue. They've also decided that no two neighboringhouses will be painted the same color. The neighbors of housei arehousesi-1 and i+1. The first and last houses are not neighbors.

You will be given the information of houses. Each house willcontain three integers"R G B" (quotes for clarity only),whereR, G and B are the costs of painting the correspondinghouse red, green, and blue, respectively. Return the minimal total costrequired 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 integern (1≤ n ≤ 20) denoting the number of houses. Each of the nextnlines will contain 3 integers "R G B". These integers will liein the range[1, 1000].

Output

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

Sample Input

Output for 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

Case 1: 137

Case 2: 96

 

题目链接：http://lightoj.com/volume_showproblem.php?problem=1047

题目大意：一共三种颜色RGB，一共n个人涂色，相邻两人不能用同一种颜色，每人每种颜色有一个花费，求最小花费

题目分析：和数字三角形类似，裸推一下就好了

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int const INF = 0x3fffffff;int const MAX = 25;int val[MAX][MAX];int dp[MAX][MAX];int main(){    int T;    scanf("%d", &T);    for(int ca = 1; ca <= T; ca++)    {        int n;        scanf("%d", &n);        for(int i = 0; i < n; i++)            for(int j = 0; j < 3; j++)                scanf("%d", &val[i][j]);        memset(dp, 0, sizeof(dp));        for(int j = 0; j < 3; j++)            dp[0][j] = val[0][j];        for(int i = 1; i < n; i++)        {            for(int j = 0; j < 3; j++)            {                if(j == 0)                    dp[i][j] = val[i][j] + min(dp[i - 1][1], dp[i - 1][2]);                else if(j == 1)                    dp[i][j] = val[i][j] + min(dp[i - 1][0], dp[i - 1][2]);                else                    dp[i][j] = val[i][j] + min(dp[i - 1][0], dp[i - 1][1]);            }        }        int ans = INF;        for(int j = 0; j < 3; j++)            ans = min(ans, dp[n - 1][j]);        printf("Case %d: %d\n", ca, ans);    }}