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点头(1163)

 

题目链接:here~~~

 

 

本题巧妙运用并查集记录每个点的前驱,更快查找出来目标状态(而且排序更加巧妙)代码某位大牛所写ORZ

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>using namespace std;pair<int,int> p[50005];int father[50005];int find(int x){    if(x <= 0) return -1;    return father[x] = (x == father[x])? x - 1 : find(father[x]);}int main(){    int n;    cin >> n;    for(int i = 0;i < n; i++){        cin>>p[i].second>>p[i].first;        p[i].first = - p[i].first;        if(p[i].second > n) p[i].second = n;    }    for(int i = 0;i <= n;i++) father[i] = i;    sort(p,p+n);    long long ans = 0;    for(int i = 0;i < n;i ++)    {        int t = find(p[i].second);        if(t >= 0) ans -= p[i].first;    }    cout<<ans<<endl;}
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点头(1163)