P1878 舞蹈课

合集 - 数据结构(28)

1.P3957 [NOIP2017 普及组] 跳房子2024-03-062.P2034 选择数字2024-03-093.P1714 切蛋糕2024-03-094.E - Insert or Erase2024-03-105.P4147 玉蟾宫2024-03-106.P2866 [USACO06NOV] Bad Hair Day S2024-03-107.P2032 扫描2024-03-108.P1884 [USACO12FEB] Overplanting S2024-03-109.P5905 【模板】全源最短路（Johnson）2024-03-1110.P1801 黑匣子2024-03-2911.P2168 [NOI2015] 荷马史诗2024-03-3012.P8306 【模板】字典树2024-03-3013.P1481 魔族密码2024-03-3014.P4341 [BJWC2010] 外星联络2024-04-0115.P1470 [USACO2.3] 最长前缀 Longest Prefix2024-04-0116.P3384 【模板】重链剖分/树链剖分2024-04-0317.P10288 [GESP样题 八级] 区间2024-04-1618.P2161 [SHOI2009] 会场预约2024-04-17

19.P1878 舞蹈课2024-05-17

20.P9691 [GDCPC2023] Base Station Construction2024-05-1821.P3611 [USACO17JAN] Cow Dance Show S2024-05-2022.F. Cutting Game2024-05-2123.G. Money Buys Less Happiness Now2024-05-2224.D. In Love2024-06-2925.F. Selling a Menagerie2024-06-2926.D. Learning to Paint2024-07-1927.P4588 [TJOI2018] 数学计算2024-07-2028.P3522 [POI2011] TEM-Temperature2024-07-21

收起

原题链接

题解

1.朴素想法：链表存储+每次遍历一遍找出最小对
缺点：时间复杂度过高
改进措施：每次遍历一遍，只会挑走一对，剩下的会重复遍历，所以我们把所有的对都找出来放进堆里，每次挑出第一个没有被用到过的对

注意审题

code

#include<bits/stdc++.h>
using namespace std;

struct node
{
    int x,y,d;
    bool operator<(const node &b)const
    {
        if(b.d!=d)return b.d<d;
        else return b.x<x;
    }
};

struct name
{
    int x,y;
};

int vis[200005]={0},a[200005],l[200005],r[200005];
vector<name> ans;
int main()
{
    int n;
    cin>>n;

    string s;
    cin>>s;
    s=' '+s;
    for(int i=1;i<=n;i++) cin>>a[i];

    priority_queue<node> q;

    for(int i=1;i<n;i++)
    {
        if(s[i]!=s[i+1])
        {
            q.push({i,i+1,abs(a[i]-a[i+1])});
        }
        l[i]=i-1;
        r[i]=i+1;
    }
    l[n]=n-1;
    r[n]=0;

    while(q.size())
    {
        int x=q.top().x,y=q.top().y;
        q.pop();
        if(vis[x]||vis[y]) continue;
        vis[x]=1;
        vis[y]=1;
        if(l[x]&&r[y]&&s[l[x]]!=s[r[y]])
            q.push({l[x],r[y],abs(a[l[x]]-a[r[y]])});
        ans.push_back({x,y});
        if(l[x]) r[l[x]]=r[y];
        if(r[y]) l[r[y]]=l[x];
    }

    cout<<ans.size()<<endl;
    for(auto it:ans)
        cout<<it.x<<" "<<it.y<<endl;
    return 0;
}