标签:style blog class code java color
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,布布扣,bubuko.com
标签:style blog class code java color
原文地址:http://www.cnblogs.com/BLADEVIL/p/3715506.html