HDU 6012 Lotus and Horticulture

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Solution

讨论版误导我好久
[li,ri]确实是闭区间，但是温度是可以取实数的，比如
1 2 100 300 200
2 3 300 100 200
ans=600 取2.5
所以我们保存坐标的时候需要保存li与ri+0.5

剩下的事情，就是排序+二分+前缀和的运用了

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define ll long long#define db doubleusing namespace std;int n,v,cas,tot;int a[50005],b[50005],c[50005];double t[100005];ll now,ans,s;struct ty{    db d;    int id;    ll sa,sb,sc;}L[50005],R[50005];bool cmp1(ty x,ty y){    return x.d>y.d;}bool cmp2(ty x,ty y){    return x.d<y.d;}int erfen1(int l,int r,db x){    if(l>r) return r;    int mid=(l+r)/2;    if(x>=L[mid].d) return erfen1(l,mid-1,x); else return erfen1(mid+1,r,x);}int erfen2(int l,int r,db x){    if(l>r) return r;    int mid=(l+r)/2;    if(x>R[mid].d) return erfen2(mid+1,r,x); else return erfen2(l,mid-1,x);}int main(){    cin>>cas;    while(cas--)     {        cin>>n;        tot=0;ans=0;s=0;        for(int i=1;i<=n;i++)         {            scanf("%lf%lf%d%d%d",&L[i].d,&R[i].d,&a[i],&b[i],&c[i]);            L[i].id=i;            R[i].id=i;            tot++;            t[tot]=L[i].d;            tot++;            t[tot]=R[i].d+0.5;            s=s+a[i];        }        sort(t+1,t+tot+1);         sort(L+1,L+n+1,cmp1);        sort(R+1,R+n+1,cmp2);        for(int i=1;i<=n;i++)         {             v=L[i].id;            L[i].sa=L[i-1].sa+a[v];            L[i].sb=L[i-1].sb+b[v];            L[i].sc=L[i-1].sc+c[v];            v=R[i].id;            R[i].sa=R[i-1].sa+a[v];            R[i].sb=R[i-1].sb+b[v];            R[i].sc=R[i-1].sc+c[v];        }        for(int i=1;i<=tot;i++)         {            v=erfen1(1,n,t[i]);            now=s-L[v].sa+L[v].sc;            v=erfen2(1,n,t[i]);            now=now-R[v].sa+R[v].sb;            ans=max(ans,now);        }        ans=max(ans,L[n].sc);        printf("%I64d\n",ans);    }    return 0;}

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