#1636 : Pangu and Stones

時間限制:1000ms 單點時限:1000ms 內存限制:256MB

描述

In Chinese mythology, Pangu is the first living being and the creator of the sky and the earth. He woke up from an egg and split the egg into two parts: the sky and the earth.

At the beginning, there was no mountain on the earth, only stones all over the land.

There were N piles of stones, numbered from 1 to N. Pangu wanted to merge all of them into one pile to build a great mountain. If the sum of stones of some piles was S, Pangu would need S seconds to pile them into one pile, and there would be S stones in the new pile.

Unfortunately, every time Pangu could only merge successive piles into one pile. And the number of piles he merged shouldn't be less than L or greater than R.

Pangu wanted to finish this as soon as possible.

Can you help him? If there was no solution, you should answer '0'.

輸入

There are multiple test cases.

The first line of each case contains three integers N,L,R as above mentioned (2<=N<=100,2<=L<=R<=N).

The second line of each case contains N integers a₁,a₂?…a_N?(1<= a_i??<=1000,i= 1…N ), indicating the number of stones of ?pile 1, pile 2 …pile N.

The number of test cases is less than 110 and there are at most 5 test cases in which N >= 50.

輸出

For each test case, you should output the minimum time(in seconds) Pangu had to take . If it was impossible for Pangu to do his job, you should output ?0.

樣例輸入

3 2 2 1 2 3 3 2 3 1 2 3 4 3 3 1 2 3 4

樣例輸出

9 6 0

【題意】

n個石子堆排成一排，每次可以將連續的最少L堆，最多R堆石子合并在一起，消耗的代價為要合并的石子總數。

求合并成1堆的最小代價，如果無法做到輸出0

?

【分析】

石子歸并系列題目，一般都是區間DP,于是——

dp[i][j][k] i到j 分為k堆的最小代價。顯然 dp[i][j][ j-i+1]代價為0

然后[i,j] 可以劃分

dp[i][j][k]? = min { dp[i][d][k-1] + dp[d+1][j][1] } (k > 1&&d-i+1 >= k-1，這個條件意思就是 區間i,d之間最少要有k-1個石子)?

最后合并的時候?

dp[i][j][1] = min{ dp[i][d][k-1] + dp[d+1][j][1]? + sum[j] - sum[i-1] }? (l<=k<=r)

?

【代碼】

#include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long ll; const int N=105; int n,L,R,s[N],f[N][N][N]; inline void Init(){for(int i=1;i<=n;i++) scanf("%d",s+i),s[i]+=s[i-1]; } inline void Solve(){memset(f,0x3f,sizeof f);for(int i=1;i<=n;i++){for(int j=i;j<=n;j++){f[i][j][j-i+1]=0;}} for(int i=n-1;i;i--){for(int j=i+1;j<=n;j++){for(int k=i;k<j;k++){for(int t=L;t<=R;t++){f[i][j][1]=min(f[i][j][1],f[i][k][t-1]+f[k+1][j][1]+s[j]-s[i-1]);}for(int t=2;t<j-i+1;t++){f[i][j][t]=min(f[i][j][t],f[i][k][t-1]+f[k+1][j][1]);}}}}ll ans=f[1][n][1];printf("%d\n",ans<0x3f3f3f3f?ans:0); } int main(){while(scanf("%d%d%d",&n,&L,&R)==3){Init();Solve();}return 0; }

轉載于:https://www.cnblogs.com/shenben/p/10494993.html

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