Description
Input
Output
Sample Input
6 12 1 3
1 9 1 2
3 2 1 2
8 20 5 4
4 11 7 4
2 10 9 1
0 0 0
Sample Output
题解:来来来,先列式子:设f[i]表示在还没有买机器i(想买还没买)时,所能得到的最大收益,那么有:
f[i]=max{f[j]+R[j]-P[j]+G[j]*(D[i]-D[j]-1)} (D[j]<D[i],f[j]>=P[j])
移项搞一搞,感觉像是斜率优化,但是好像x不单调?依旧套用cash那题的做法。上cdq分治,左边按x单调,右边按k单调,用左边更新右边即可。
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; typedef long long ll; const int maxn=100010; int n,h,t,T; int q[maxn]; struct node { ll P,R,D,G,f; }s[maxn],p[maxn]; ll X(int a) { return s[a].G; } ll Y(int a) { return s[a].P-s[a].R+s[a].G+s[a].G*s[a].D-s[a].f; } int rd() { int ret=0,f=1; char gc=getchar(); while(gc<'0'||gc>'9') {if(gc=='-')f=-f; gc=getchar();} while(gc>='0'&&gc<='9') ret=ret*10+gc-'0',gc=getchar(); return ret*f; } bool cmpx(node a,node b) { return a.G<b.G; } bool cmpk(node a,node b) { return a.D<b.D; } long double slope(int a,int b) { if(X(a)==X(b)) return (long double)(Y(a)<Y(b)?2147483647:-2147483647); return (long double)(Y(b)-Y(a))/(X(b)-X(a)); } void solve(int l,int r) { if(l==r) { s[l].f=max(s[l].f,s[l-1].f); return ; } int mid=l+r>>1,i,h1=l,h2=mid+1; solve(l,mid); h=1,t=0; for(i=l;i<=mid;i++) { if(s[i].f<s[i].P) continue; while(h<t&&slope(q[t-1],q[t])>=slope(q[t],i)) t--; q[++t]=i; } for(i=mid+1;i<=r;i++) { while(h<t&&slope(q[h],q[h+1])<=s[i].D) h++; s[i].f=max(s[i].f,s[i-1].f); if(h<=t) s[i].f=max(s[i].f,s[q[h]].f-s[q[h]].P+s[q[h]].R+s[q[h]].G*(s[i].D-s[q[h]].D-1)); } solve(mid+1,r); for(i=l;i<=r;i++) { if(h1<=mid&&(h2>r||X(h1)<=X(h2))) p[i]=s[h1++]; else p[i]=s[h2++]; } for(i=l;i<=r;i++) s[i]=p[i]; } int main() { while(1) { memset(s,0,sizeof(s)); n=rd(),s[0].f=rd(),s[n+1].D=rd()+1; if(!n) return 0; int i; for(i=1;i<=n;i++) s[i].D=rd(),s[i].P=rd(),s[i].R=rd(),s[i].G=rd(); n++; sort(s,s+n+1,cmpk); solve(0,n); sort(s,s+n+1,cmpk); printf("Case %d: %lld\n",++T,s[n].f); } }
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