ACM学习笔记DAY2

今天学习的主要内容为模拟与贪心

• 模拟
  就是根据实际问题建立模型，模拟实际程序走一遍。（但是有些思维的难度较大）
  例如：

#include<iostream>
#include<stdio.h>
#pragma warning(disable:4996)
#define ll long long
int main()
{ll n;while (~scanf("%lld", &n)){for (ll i = 2; i * i <= n; i++)while (n % (i * i) == 0)n /= i;printf("%lld\n", n);}return 0;
}

• 贪心
  贪心算法是指不从整体上最优解考虑，算法得到的是某种意义上的最优解。
  

这道题有两种方案，一种是让速度最快的人往返接送，最快的人来来往往送完全部的人过河，每次的往返为一个局部，时间为t1
第二种方案是最快的人与次快的人去对岸，最快的人回到此岸，最慢的人再和次慢的人去对岸，次快的人回到此案此岸，时间为t2

之后最优的解就是min（t1,t2）

#include<iostream>
#include<algorithm>
using namespace std;
int a[10000];
int main() {int t;cin >> t;while (t--){int sum = 0;int n;cin >> n;for (int i = 0; i < n; i++){cin >> a[i];}sort(a , a + n );while (n > 3) { int t1 = 2 * a[0] + a[n - 1] + a[n - 2];int t2 = 2 * a[1] + a[0] + a[n - 1];sum += min(t1, t2);n -= 2;}if (n == 1 || n == 3) {for (int i = 0; i < n; i++) sum += a[i];}else if (n == 2) sum += a[1];cout << sum << endl;}
}

接下来是今天在vj上遇到的没有ac的题

这道题当时写的时候一直超时，最后在考试结束后看了题解才ac

#include<iostream>
#include <algorithm>
#pragma warning(disable:4996)
using namespace std;
#define maxn 25005struct Node {int u, v;
} E[maxn];bool cmp(Node a, Node b) {return a.u < b.u;
}int main() {int N, T, flag, ans, right, i;while (scanf("%d%d", &N, &T) == 2) {for (i = 0; i < N; ++i)scanf("%d%d", &E[i].u, &E[i].v);sort(E, E + N, cmp);ans = right = 0;flag = 1;i = 0;while (flag <= T) {for (; i < N && E[i].u <= flag; ++i)if (E[i].u <= flag && E[i].v > right)right = E[i].v;if (right >= flag) flag = right + 1, ++ans;else break;}if (flag <= T) ans = -1;printf("%d\n", ans);}return 0;
}

至于下面这一道题

当时看完题目头都大了，测试样例都没有过。。。所以留下大佬的ac代码与大家共同学习。

#include <bits/stdc++.h>
#define ll long long
#define time mytime
using namespace std;const int N = 1e5;set<pair<ll,int> > set1,set2;
int n,m;
ll bounty,increase,damage;
ll max_h[N+10],start_h[N+10],regen[N+10];
ll time[N+10],enemy[N+10],health[N+10],dic[N+10],delta = 0;
bool special[N+10];int main(){ios::sync_with_stdio(0),cin.tie(0);memset(dic,255,sizeof dic);cin >> n >> m;cin >> bounty >> increase >> damage;for (int i = 1;i <= n;i++){cin >> max_h[i] >> start_h[i] >> regen[i];if (start_h[i]>damage) continue;if (max_h[i]<=damage ||(start_h[i]<=damage && regen[i]==0)){special[i] = 1;delta++;continue;}ll time = (damage-start_h[i])/regen[i];dic[i] = time;set1.insert({time,i});}for (int i = 1;i <= m;i++){cin >> time[i] >> enemy[i] >> health[i];set2.insert({time[i],i});}ll ans = 0;int pre = -1;ll prenum = 0;while (!set1.empty() || !set2.empty()){ll t1 = -1,t2 = -1;if (!set1.empty()) t1 = (*set1.begin()).first;if (!set2.empty()) t2 = (*set2.begin()).first;//等于的话先处理更新的if (t1!=-1 &&(t2==-1 || t1 <t2)){ll time2 = (*set1.begin()).first;ll num = (ll)set1.size();if (pre!=-1 && time2==pre){num = prenum;}else {pre = time2;prenum = num;}ans = max(ans,(num+delta)*(bounty+t1*increase));set1.erase(set1.begin());}else{ll num = (ll)set1.size();ans = max(ans,(num+delta)*(bounty+(t2-1)*increase));int tempidx = (*set2.begin()).second;set2.erase(set2.begin());int idx =enemy[tempidx];auto it = set1.upper_bound({dic[idx],idx});if (it!=set1.begin()){it--;pair<ll,int> temp = (*it);if (temp.first==dic[idx] && temp.second==idx){set1.erase(it);}}if (special[idx]) delta--;special[idx] = 0;dic[idx] = -1;if (health[tempidx]>damage) continue;if (max_h[idx]<=damage ||(health[tempidx]<=damage && regen[idx]==0)){special[idx] = 1;delta++;continue;} else{ll time1 = (damage-health[tempidx])/regen[idx];dic[idx] = time[tempidx] + time1;set1.insert({dic[idx],idx});}}}int cc =0;for (int i = 1;i <= n;i++)if (special[i])cc++;if (cc>0 && increase>0)return cout<<-1,0;else ans = max(ans,(ll)cc*bounty);cout << ans << endl;return 0;
}

-7.7 共勉