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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7574 Accepted Submission(s): 1667
题解：
http://www.docin.com/p-47950655.html这篇论文讲的斜率优化，讲的很清楚；
总结就是：构造下凸折线，维护下凸折线（凸包维护），找与下凸折线相切的斜率（也可以用二分来找）
没用二分：
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<vector> using namespace std; const int INF=0x3f3f3f3f; #define mem(x,y) memset(x,y,sizeof(x)) #define SI(x) scanf("%d",&x) #define PI(x) printf("%d",x) #define SD(x) scanf("%lf",&x) #define P_ printf(" ") typedef long long LL; const int MAXN=1e5+100; int sum[MAXN],s[MAXN],a[MAXN]; bool cross(int i,int j,int k){ if((sum[j]-sum[i])*(k-i)>=(sum[k]-sum[i])*(j-i))return true; return false; } double flx(int i,int t){ double temp; temp=1.0*(sum[t]-sum[i])/(t-i); return temp; } int main(){ int N,k; while(~scanf("%d%d",&N,&k)){ sum[0]=0; for(int i=1;i<=N;i++)SI(a[i]),sum[i]=sum[i-1]+a[i]; int top=0,low=0; double ans=0; for(int i=k;i<=N;i++){ int j=i-k; while(top-low>=1&&cross(s[top-1],s[top],j))top--; s[++top]=j; while(top-low>=1&&flx(s[low+1],i)>=flx(s[low],i))low++; ans=max(ans,flx(s[low],i)); } printf("%.2lf\n",ans); } return 0; }
用了二分：
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<vector> using namespace std; const int INF=0x3f3f3f3f; #define mem(x,y) memset(x,y,sizeof(x)) #define SI(x) scanf("%d",&x) #define PI(x) printf("%d",x) #define SD(x) scanf("%lf",&x) #define P_ printf(" ") typedef long long LL; const int MAXN=1e5+100; int sum[MAXN],s[MAXN],a[MAXN]; bool cross(int i,int j,int k){ if((sum[j]-sum[i])*(k-i)>=(sum[k]-sum[i])*(j-i))return true; return false; } double flx(int i,int t){ double temp; temp=1.0*(sum[t]-sum[i])/(t-i); return temp; } int erfen(int l,int r,int i){ int mid; while(l<=r){ mid=(l+r)>>1; if(cross(s[mid],s[mid+1],i))r=mid-1; else l=mid+1; } return r+1; } int main(){ int N,k; while(~scanf("%d%d",&N,&k)){ sum[0]=0; for(int i=1;i<=N;i++)SI(a[i]),sum[i]=sum[i-1]+a[i]; int top=0,low=0; double ans=0; for(int i=k;i<=N;i++){ int j=i-k; while(top-low>=1&&cross(s[top-1],s[top],j))top--; s[++top]=j; ans=max(ans,flx(s[erfen(0,top,i)],i)); } printf("%.2lf\n",ans); } return 0; }
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原文地址：http://www.cnblogs.com/handsomecui/p/5205831.html
