HDU 3507 Print Article（CDQ分治+分治DP）
forever97
posted @ 2016年9月20日 18:21
in 算法-分治DP
with tags
CDQ分治，分治DP
, 765 阅读
【题目链接】 http://acm.hdu.edu.cn/showproblem.php?pid=3507
【题目大意】
将长度为n的数列分段，最小化每段和的平方和。
【题解】
根据题目很容易得到dp[j]=min(dp[k]+(s[j]-s[k])2)，因为是从前往后转移，且决策单调，因此在CDQ分治的同时进行分治DP即可。
【代码】
#include <cstdio>
typedef long long LL;
const int N=500005;
int n,M,t;
LL f[N],g[N],a[N],s[N],INF=1LL<<60;
void DP(int l,int r,int dl,int dr){
int m=(l+r)>>1,i,dm=0;
LL &ret=g[m]; ret=INF;
for(i=dl;i<=dr&&i<m;i++){
LL t=f[i]+(s[m]-s[i])*(s[m]-s[i])+M;
if(t<ret)ret=t,dm=i;
}if(l<m)DP(l,m-1,dl,dm);
if(r>m)DP(m+1,r,dm,dr);
}
void CDQ(int l,int r){
if(l==r)return;
int mid=(l+r)>>1;
CDQ(l,mid);
DP(mid+1,r,l,mid);
for(int i=r;i>mid;i--)if(g[i]<f[i])f[i]=g[i];
CDQ(mid+1,r);
}
int main(){
while(~scanf("%d%d",&n,&M)){
for(int i=1;i<=n;i++){scanf("%lld",&s[i]);s[i]+=s[i-1];}
for(int i=1;i<=n;i++)f[i]=s[i]*s[i]+M;CDQ(0,n);
printf("%lld\n",f[n]);
}return 0;
}
2024年1月18日 16:41
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