NOIP2012 T3开车旅行 set+倍增

70分做法:
先预处理出所有点的最近和次近(O(n^2)一遍就OK)
然后暴力求出每个解(O(nm))

//By SiriusRen
#include <cstdio>
#include <cstring>
#include <algorithm>
#define inf 0x3fffffff
using namespace std;
int n,x,rech=0x3fffffff,rec,s,m;
double ans=0x3fffffff;
struct Path{int to,weight,to2,weight2;void init(){weight=inf,weight2=inf,to=-1,to2=-1;}}path[100050];
struct Node{int position,height;}node[100050];
bool cmp(Node a,Node b){return a.height<b.height;}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        node[i].position=i;
        scanf("%d",&node[i].height);
    }
    for(int i=1;i<=n;i++){
        path[i].init();
        for(int j=i+1;j<=n;j++){
            if(path[i].weight2>abs(node[j].height-node[i].height)){
                path[i].weight2=abs(node[j].height-node[i].height);
                path[i].to2=node[j].position;
            }
            else if(path[i].weight2==abs(node[j].height-node[i].height)&&node[j].height<node[i].height){
                path[i].to2=node[j].position;
            }
            if(path[i].weight2<path[i].weight){
                swap(path[i].weight2,path[i].weight);
                swap(path[i].to,path[i].to2);
            }
            else if(path[i].weight2==path[i].weight&&node[j].height<node[i].height){
                swap(path[i].to,path[i].to2);
            }
        }
    }
    scanf("%d",&x);
    for(int i=1;i<=n;i++){
        int wei1=0,wei2=0,f=1;
        for(int j=i;;){
            if(f){
                if(wei1+wei2+path[j].weight2<=x)
                    wei2+=path[j].weight2;
                else break;
                j=path[j].to2;
            }
            else{
                if(wei1+wei2+path[j].weight<=x)
                    wei1+=path[j].weight;
                else break;
                j=path[j].to;
            }
            f^=1;
        }
        if(wei1&&(ans>1.0*wei2/wei1||(ans==1.0*wei2/wei1&&rech<node[i].height))){
            ans=1.0*wei2/wei1;
            rec=i;
            rech=node[i].height;
        }
    }
    printf("%d
",rec);
    scanf("%d",&m);
    for(int i=1;i<=m;i++){
        scanf("%d%d",&s,&x);
        int wei1=0,wei2=0,f=1;
        for(int j=s;;){
            if(f){
                if(wei1+wei2+path[j].weight2<=x)
                    wei2+=path[j].weight2;
                else break;
                j=path[j].to2;
            }
            else{
                if(wei1+wei2+path[j].weight<=x)
                    wei1+=path[j].weight;
                else break;
                j=path[j].to;
            }
            f^=1;
        }
        printf("%d %d
",wei2,wei1); 
    }
}

100分做法:
先用set 从后向前插入 ,取出左边两个点和右边两个点(如果有的话),排个序。
O(nlogn)求出最近和次近。
然后呢 用倍增求距离
g[i][j]表示从i出发走2^j轮到的地方
f[i][j][0]表示从i出发走2^j轮A走了多少
f[i][j][1]表示从i出发走2^j轮B走了多少
预处理出走一轮到哪儿
g[i][0]=edge[edge[i].to2].to;
f[i][0][0]=edge[i].weight2;
f[i][0][1]=edge[edge[i].to2].weight;

倍增就好了…
g[i][j]=g[g[i][j-1]][j-1];
f[i][j][0]=f[g[i][j-1]][j-1][0]+f[i][j-1][0];
f[i][j][1]=f[g[i][j-1]][j-1][1]+f[i][j-1][1];

//By SiriusRen
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define int long long
using namespace std;
int n,xx,x,m,g[100050][20],f[100050][20][2],rech,rec;
double ans=0x3fffffff;
struct Node{int height,position;}node[100050];
struct Node2{int height,Short,position;}jy;
struct Edge{int to,to2,weight,weight2;void init(){weight=weight2=0x3fffffff;}}edge[100050];
set<Node>s;set<Node2>p;
bool operator < (Node a,Node b){return a.height<b.height;}
bool operator < (Node2 a,Node2 b){
    if(a.Short!=b.Short)return a.Short<b.Short;
    else return a.height<b.height;
}
void init(){
    for(int i=n;i>=1;i--){
        p.clear();edge[i].init();
        s.insert(node[i]);
        set<Node>::iterator it=s.find(node[i]),it2=it;
        if((++it)!=s.end()){
            jy.height=(*it).height;
            jy.position=(*it).position;
            jy.Short=abs((*it).height-node[i].height);
            p.insert(jy);
        }
        else goto deal1;
        if((++it)!=s.end()){
            jy.height=(*it).height;
            jy.position=(*it).position;
            jy.Short=abs((*it).height-node[i].height);
            p.insert(jy);
        }
        deal1:if(it2==s.begin())goto deal;
        if((--it2)==s.begin()){
            jy.height=(*it2).height;
            jy.position=(*it2).position;
            jy.Short=abs((*it2).height-node[i].height);
            p.insert(jy);
            goto deal;
        }
        else
        {
            jy.height=(*it2).height;
            jy.position=(*it2).position;
            jy.Short=abs((*it2).height-node[i].height);
            p.insert(jy);
        }
        it2--;
        jy.height=(*it2).height;
        jy.position=(*it2).position;
        jy.Short=abs((*it2).height-node[i].height);
        p.insert(jy);
        deal:set<Node2>::iterator itp=p.begin();
        if(itp!=p.end()){
            edge[i].weight=(*itp).Short;
            edge[i].to=(*itp).position;
            if((++itp)!=p.end()){
                edge[i].weight2=(*itp).Short;
                edge[i].to2=(*itp).position;
            }
        }
    }
}
signed main(){
    scanf("%lld",&n);
    for(int i=1;i<=n;i++){
        scanf("%lld",&node[i].height);
        node[i].position=i;
    }
    init();
    for(int i=1;i<=n;i++){
        g[i][0]=edge[edge[i].to2].to;
        f[i][0][0]=edge[i].weight2;
        f[i][0][1]=edge[edge[i].to2].weight;
    }
    for(int j=1;j<=18;j++){
        for(int i=1;i<=n;i++){
            g[i][j]=g[g[i][j-1]][j-1];
            f[i][j][0]=f[g[i][j-1]][j-1][0]+f[i][j-1][0];
            f[i][j][1]=f[g[i][j-1]][j-1][1]+f[i][j-1][1];
        }
    }
    scanf("%lld%lld",&x,&m);
    for(int i=1;i<=n;i++){
        int wei1=0,wei2=0,S=i,temp=x;
        for(int j=17;j>=0;j--){
            if(temp>=f[S][j][0]+f[S][j][1]&&g[S][j]){
                wei1+=f[S][j][0];
                wei2+=f[S][j][1];
                temp=temp-(f[S][j][0]+f[S][j][1]);
                S=g[S][j];
            }
        }
        if(temp>=edge[S].weight2&&edge[S].to2){
            temp-=edge[S].weight2;
            wei1+=edge[S].weight2;
        }
        if(wei2&&(ans>1.0*wei1/wei2||(ans==1.0*wei1/wei2&&rech<node[i].height))){
            rech=node[i].height;
            rec=i;
            ans=1.0*wei1/wei2;
        }
    }
    printf("%lld
",rec);
    for(int i=1;i<=m;i++){
        scanf("%lld%lld",&xx,&x);
        int wei1=0,wei2=0,S=xx,temp=x;
        for(int j=17;j>=0;j--){
            if(temp>=f[S][j][0]+f[S][j][1]&&g[S][j]){
                wei1+=f[S][j][0];
                wei2+=f[S][j][1];
                temp=temp-(f[S][j][0]+f[S][j][1]);
                S=g[S][j];
            }
        }
        if(temp>=edge[S].weight2&&edge[S].to2){
            temp-=edge[S].weight2;
            wei1+=edge[S].weight2;
        }
        printf("%lld %lld
",wei1,wei2);
    }
}

这里写图片描述

原文地址:https://www.cnblogs.com/SiriusRen/p/6532309.html