【lightoj1047】Neighbor House
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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 arehouses i-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),where R, 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 next nlines 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
dp题目,看来还是对dp理解不够深啊,没看出来用dp。其实看出来dp后算是一道dp的水题。
#include<cstdio>#include<iostream>#include<cstring>#include<cmath>#define INF 0x3f3f3f3f#include<algorithm>using namespace std;int map[22][5],dp[22][5];int main() {int T,p=0;scanf("%d",&T);while(T--) {//printf("\n"); //加上格式错误。int n;memset(dp,0,sizeof(dp));scanf("%d",&n);for(int i=1; i<=n; i++) {for(int j=0; j<3; j++) {scanf("%d",&map[i][j]);}}int ans=0,tab=4;dp[1][0]=map[1][0];dp[1][1]=map[1][1];dp[1][2]=map[1][2];for(int i=2; i<=n; i++) {dp[i][0]=min(dp[i-1][1]+map[i][0],dp[i-1][2]+map[i][0]);dp[i][1]=min(dp[i-1][0]+map[i][1],dp[i-1][2]+map[i][1]);dp[i][2]=min(dp[i-1][0]+map[i][2],dp[i-1][1]+map[i][2]);}ans=min(min(dp[n][0],dp[n][1]),dp[n][2]);printf("Case %d: %d\n",++p,ans);}return 0;}
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