标签:oid 复杂 ace space fine sum bre php 复杂度
HDU 3507:Print Article
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3507
题目大意:给定$n$,$m$,输出序列$n$个数,每连续输出代价为连续输出的数字和的平方加上$m$.
斜率优化DP
定义$sum_{pq}=\sum_{k=p+1}^qa_k=pre[q]-pre[p]$,其中$pre[]$维护的是前缀和.
直接DP:$dp[i]=min\{dp[q]+sum_{qi}+m\}(q<i)$,如此复杂度为$O(n^2)$,故需要优化.
设$p<q$,若从$q$转移而来比从$p$更优,则有$dp[q]+sum_{qi}+m<dp[p]+sum_{pi}+m$,
化简可得,$\frac{(dp[q]+pre[q]^2)-(dp[p]+pre[p]^2)}{2pre[q]-2pre[p]}<pre[i]$.
令$y(k)=dp[k]+pre[k]^2$,$x(k)=2pre[k]$,
则上式可写为$\frac{y(q)-y(p)}{x(q)-x(p)}<pre[i]$.
其中,$\frac{y(q)-y(p)}{x(q)-x(p)}$可看做斜率,定义$g(p,q)=\frac{y(q)-y(p)}{x(q)-x(p)}$,
故有,若$g(p,q)<pre[i]$,则$q$比$p$更优.
待更...
代码如下:
1 #include <cstdio> 2 #define N 500005 3 using namespace std; 4 typedef long long ll; 5 int n,m,dp[N],a[N],pre[N],deq[N],r,f; 6 int y(int n){return dp[n]+pre[n]*pre[n];} 7 int x(int n){return 2*pre[n];} 8 int main(void){ 9 while(~scanf("%d%d",&n,&m)){ 10 r=-1,f=0; 11 deq[++r]=0; 12 for(int i=1;i<=n;++i){ 13 scanf("%d",&a[i]); 14 pre[i]=pre[i-1]+a[i]; 15 } 16 for(int i=1;i<=n;++i){ 17 while(r-f>0){ 18 if(y(deq[f+1])-y(deq[f]) 19 <=pre[i]*(x(deq[f+1])-x(deq[f]))) 20 f++; 21 else break; 22 } 23 dp[i]=dp[deq[f]]+(pre[i]-pre[deq[f]])*(pre[i]-pre[deq[f]])+m; 24 while(r-f>0){ 25 if((y(deq[r])-y(deq[r-1]))*(x(i)-x(deq[r])) 26 >=(y(i)-y(deq[r]))*(x(deq[r])-x(deq[r-1]))) 27 r--; 28 else break; 29 } 30 deq[++r]=i; 31 } 32 printf("%d\n",dp[n]); 33 } 34 }
标签:oid 复杂 ace space fine sum bre php 复杂度
原文地址:http://www.cnblogs.com/barrier/p/6816390.html