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HDU3507 Print Article

2024-07-20 17:33:21   221人阅读

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Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)


Problem Description
Zero has an old printer that doesn‘t work well sometimes. As it is antique, he still like to use it to print articles. But it is too old to work for a long time and it will certainly wear and tear, so Zero use a cost to evaluate this degree.
One day Zero want to print an article which has N words, and each word i has a cost Ci to be printed. Also, Zero know that print k words in one line will cost

M is a const number.
Now Zero want to know the minimum cost in order to arrange the article perfectly.
 

 

Input
There are many test cases. For each test case, There are two numbers N and M in the first line (0 ≤ n ≤ 500000, 0 ≤ M ≤ 1000). Then, there are N numbers in the next 2 to N + 1 lines. Input are terminated by EOF.
 

 

Output
A single number, meaning the mininum cost to print the article.
 

 

Sample Input
5 559575
 

 

Sample Output
230
 

 

Author
Xnozero
 

 

Source
2010 ACM-ICPC Multi-University Training Contest（7）——Host by HIT
 

 

Recommend
zhengfeng
 
裸的斜率dp。。注意不等号方向。。样例数据挺好的。。不管你是什么符号都输出230..23333
 1 #include<cstdio> 2 #include<cstdlib> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 const int N = 500100; 7 #define For(i,n) for(int i=1;i<=n;i++) 8 #define Rep(i,l,r) for(int i=l;i<=r;i++) 9 10 int n,M,sum[N],dp[N],q[N];11 //  dp[i] = max{dp[j] + (sj-si)^2 + M}12 //-> dp[k] - dp[j] + sk^2 - sj^2 13 //    --------------------------   > 2si 时 j比k优 14 //              sk - sj15 16 int Up(int i,int j){17     return dp[i]+sum[i]*sum[i]-(dp[j]+sum[j]*sum[j]);18 }19 20 int Down(int i,int j){21     return sum[i]-sum[j]; 22 }23 24 void DP(){25     int l=0,r=1;26     For(i,n){27         while(l+1<r && Up(q[l],q[l+1])>=2*sum[i]*Down(q[l],q[l+1])) l++;28         dp[i]=dp[q[l]] + (sum[i]-sum[q[l]])*(sum[i]-sum[q[l]]) + M;29         while(l+1<r && Up(q[r-2],q[r-1])*Down(q[r-1],i)>=Up(q[r-1],i)*Down(q[r-2],q[r-1])) --r;30         q[r++]=i;31     } 32     printf("%d\n",dp[n]);33 }34 35 int main(){36     while(scanf("%d%d",&n,&M)!=EOF){37         q[0]=dp[0]=0;38         For(i,n){39             scanf("%d",&sum[i]);40             sum[i]+=sum[i-1];41         }42         DP();43     }44     return 0;45 }

 

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