Colorful Slimes
时间限制: 2 Sec 内存限制: 256 MB 提交: 235 解决: 36 [ ][ ][ ][命题人: ]题目描述
Snuke lives in another world, where slimes are real creatures and kept by some people. Slimes come in N colors. Those colors are conveniently numbered 1 through N. Snuke currently has no slime. His objective is to have slimes of all the colors together. Snuke can perform the following two actions: Select a color i (1≤i≤N), such that he does not currently have a slime in color i, and catch a slime in color i. This action takes him ai seconds. Cast a spell, which changes the color of all the slimes that he currently has. The color of a slime in color i (1≤i≤N−1) will become color i+1, and the color of a slime in color N will become color 1. This action takes him x seconds. Find the minimum time that Snuke needs to have slimes in all N colors. Constraints 2≤N≤2,000 ai are integers. 1≤ai≤109 x is an integer. 1≤x≤109
输入
The input is given from Standard Input in the following format: N x a1 a2 … aN
输出
Find the minimum time that Snuke needs to have slimes in all N colors.
样例输入
2 10
1 100
样例输出
12
提示
Snuke can act as follows:
Catch a slime in color 1. This takes 1 second.Cast the spell. The color of the slime changes: 1 → 2. This takes 10 seconds.Catch a slime in color 1. This takes 1 second.来源
【题意】
要获得n中颜色， 有两种操作
一 直接制造 花费时间a[i]秒
二 现拥有的全部转换 i-1 -> i ,i -> i+1, i+1 -> i+2 花费时间x 秒
问获得n 种颜色 需要 最少时间；
dp[i][j] i 代表制作第i 个颜色， j 代表转换次数；
最优 应该为 k*x + sum( dp[i][k])
先dp 每种颜色 最优的策略 ， 然后 计算 k中 转换里 sum 和 最小的。 加上 k*x
因为 是 转换一次， 那么 全部都需要 转换，所有 n 中颜色 实际上 是 一体的。
【code】
#includeusing namespace std;const int N=2000+10;typedef long long ll;const ll llf=(1ll<<63)-1;ll dp[N][N],a[N];int main(){ int n; ll x,ans=llf; scanf("%d%lld",&n,&x); for(int i=1;i<=n;i++) { scanf("%lld",&a[i]); dp[0][i]=a[i]; } for(int i=1;i<=n-1;i++) { for(int j=1;j<=n;j++) { int ind=j-i; if(ind<=0) ind=n+ind; dp[i][j]=min(dp[i-1][j],a[ind]); } } for(int i=0;i<=n-1;i++) { ll ant=0; for(int j=1;j<=n;j++) ant+=dp[i][j]; ans=min(ans,ant+i*x); // cout< <
123