light oj1074
2015-08-16 18:53
337 查看
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
Hint
Use simple DP
题意:
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
Hint
Use simple DP
题意:
有n户人,打算把他们的房子图上颜色,有red、green、blue三种颜色,每家人涂不同的颜色要花不同的费用,
而且相邻两户人家之间的颜色要不同,求最小的总花费费用。
思路:
这个题与刘汝佳的算法竞赛与入门经典中的数字三角形有点类似,可以参照其方法,利用动态规划的思想,找出
状态转移方程,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],这样答案就出来了。
代码:
#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]);
}
}
printf("Case %d: %d\n", k, min(dp
[1], min(dp
[2], dp
[3])));
}
return 0;
}
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
Hint
Use simple DP
题意:
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
Hint
Use simple DP
题意:
有n户人,打算把他们的房子图上颜色,有red、green、blue三种颜色,每家人涂不同的颜色要花不同的费用,
而且相邻两户人家之间的颜色要不同,求最小的总花费费用。
思路:
这个题与刘汝佳的算法竞赛与入门经典中的数字三角形有点类似,可以参照其方法,利用动态规划的思想,找出
状态转移方程,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],这样答案就出来了。
代码:
#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]);
}
}
printf("Case %d: %d\n", k, min(dp
[1], min(dp
[2], dp
[3])));
}
return 0;
}
相关文章推荐
- [人月神话]读书笔记6--永恒的变化&&开发工具效用
- HDU-5344
- 我的linux分区方案
- 【转】如何给Ubuntu添加Windows及Mac字体?
- join sleep yield
- 如何解决ubuntu下Chromium 新建的应用快捷方式图标模糊的问题
- linux下的输出重定向和快捷键
- 字符串匹配算法---KMP算法
- 【转】解决Ubuntu下Sublime Text 3无法输入中文
- Java中E、T、K、V、N的含义
- Java神奇的装箱与拆箱
- 解决ubuntu中JDK的Picked up JAVA_TOOL_OPTIONS提示问题。
- vs2015正式版,建立安卓工程报错:值不能为空,参数名:path1的错误解决
- 【转】eclipse下使用git上传(下载)代码至(从)github
- 我的linux分区方案
- 【转】如何给Ubuntu添加Windows及Mac字体?
- zoj 3537 Cake (凸包确定+间隔dp)
- win8.1下修复ieframe.dll的一种办法
- hdu 5072 Coprime (容斥)
- 【转载】中文ubuntu里用户目录里的路径改成英文