Light OJ 1047 Neighbor House

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 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	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;

const int INF = 0x3fffffff;

int t, ca, n, val[25][25], dp[25][25];
int main()
{
    scanf("%d", &t);
    for (ca = 1; ca <= t; ca++){
        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);
    }
    return 0;
}

#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std;

int dp[23][4];
int a[23][4];

int main()
{
    int n, t;
    scanf("%d",&t);
    for(int k = 1; k <= t; k++)
    {
        scanf("%d",&n);
        memset(a, 0, sizeof(a));
        memset(dp, 0, sizeof(dp));
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= 3; j++)
                scanf("%d",&a[i][j]);
        }
        for (int i = 1; i <= n; i++)
        {
            for (int j = 3; j < 6; j++)
            {
                dp[i][j%3+1] = a[i][j%3+1] + min(dp[i-1][(j-1)%3+1], dp[i-1][(j+1)%3+1]);
                //这里我用这种方法避免了分情况讨论，、减少了代码量
            }
        }
        printf("Case %d: %d\n", k, min(dp[n][1], min(dp[n][2], dp[n][3])));
    }
    return 0;
}