bzoj 3437 斜率优化DP

　　写题解之前首先要感谢妹子。

　　比较容易的斜率DP，设sum[i]=Σb[j]，sum_[i]=Σb[j]*j，w[i]为第i个建立，前i个的代价。

　　那么就可以转移了。

　　备注：还是要感谢妹子。

/**************************************************************

    Problem: 3437

    User: BLADEVIL

    Language: C++

    Result: Accepted

    Time:3404 ms

    Memory:39872 kb

****************************************************************/

//By BLADEVIL

#include <cstdio>

#define maxn 1000010

#define LL long long
using namespace std;
LL n;

LL a[maxn],sum[maxn],sum_[maxn];

LL que[maxn];

LL w[maxn];
LL get(LL i) {

    return (w[i]+sum_[i]);

}
int main() {

    scanf("%lld",&n);

    for (LL i=1;i<=n;i++) scanf("%lld",&a[i]);

    for (LL i=1;i<=n;i++) scanf("%lld",&sum[i]);

    for (LL i=1;i<=n;i++) sum_[i]=i*sum[i];

    for (LL i=1;i<=n;i++) sum[i]+=sum[i-1],sum_[i]+=sum_[i-1];

    LL h(1),t(1);

    for (LL i=1;i<=n;i++) {

        while ((t-h>0)&&((sum[que[h]]-sum[que[h+1]])*i<=(w[que[h]]+sum_[que[h]]-w[que[h+1]]-sum_[que[h+1]]))) h++;

        w[i]=w[que[h]]+(sum[i]-sum[que[h]])*i-(sum_[i]-sum_[que[h]])+a[i];

        //w[i]=w[que[h]]+sum_[que[h]]-i*sum[que[h]]+i*sum[i]+a[i]-sum_[i];

        while ((t>h)&&((get(que[t-1])-get(i))*(sum[que[t-1]]-sum[que[t]])<=(get(que[t-1])-get(que[t]))*(sum[que[t-1]]-sum[i]))) t--;

        /*

        while ((t-h>0)&&(

        (w[que[t-1]]+sum_[que[t-1]]-w[i]-sum_[i])*(sum[que[t-1]]-sum[que[t]])<=

                         (w[que[t-1]]+sum_[que[t-1]]-w[que[t]]-sum_[que[t]])*(sum[que[t-1]]-sum[i]))

        ) t--;

        */

        que[++t]=i;

    }

    printf("%lld\n",w[n]);

    return 0;

}

bzoj 3437 斜率优化DP