洛谷刷题7.25

P1102 A-B 数对 - 洛谷

 该题可用尺取法，二分法。也可用map比较省事，nlogn的复杂度可以接受

#include<bits/stdc++.h>
#define ll long long
using namespace std;
map<ll,ll>q; 
ll n,c,arr[200005],ans=0;
int main(){
cin>>n>>c;
for(int i=1;i<=n;i++){cin>>arr[i];q[arr[i]]++;
}
for(int i=1;i<=n;i++){if(q.find(arr[i]+c)!=q.end()) ans+=q[arr[i]+c];
}
cout<<ans;return 0;
}

P1462 通往奥格瑞玛的道路 - 洛谷

 二分+dijsktra求最短路

#include<bits/stdc++.h>
#define ll long long
using namespace std;
struct bian{ll u,l;bool operator<(const bian &s1)const{return s1.l<l;}
};
ll n,m,k,f[10005],a,b,c,s[10005],d[10005];
bool vis[10005];
vector<bian>arr[10005];
bool check(ll x){memset(d,0x3f,sizeof(d));memset(vis,false,sizeof(vis));d[1]=0;priority_queue<bian>r; if(f[1]<=x) r.push(bian{1,0});while(!r.empty()){a=r.top().u;if(!vis[a]){vis[a]=true;for(auto it:arr[a]){if(f[it.u]<=x){if(d[it.u]>d[a]+it.l){d[it.u]=d[a]+it.l;r.push(bian{it.u,d[it.u]});	}}}}r.pop();}if(d[n]<=k) return true;else return false;
}
int main(){
cin>>n>>m>>k;
for(int i=1;i<=n;i++){cin>>f[i];s[i]=f[i];
}
sort(s+1,s+1+n);
for(int i=0;i<m;i++){cin>>a>>b>>c;arr[a].push_back(bian{b,c});arr[b].push_back(bian{a,c});
}
int l=0,r=n+1;
while(l+1<r){int mid=(l+r)/2;if(check(s[mid])) r=mid;else l=mid;
}
if(r==n+1) cout<<"AFK";
else cout<<s[r];return 0;
}

 P1824 [USACO05FEB] 进击的奶牛 Aggressive Cows G - 洛谷

 依旧二分答案

#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll n,m,arr[100005],sum;
bool check(ll x){ll now=arr[1];sum=1;for(int i=2;i<=n;i++){if(arr[i]-now>=x){sum++;now=arr[i];}if(sum>=m) return true;}return false;
} 
int main(){
cin>>n>>m;
for(int i=1;i<=n;i++){cin>>arr[i];
}
sort(arr+1,arr+1+n);
ll l=0,r=1e9;
while(l+1<r){ll mid=(l+r)/2;if(check(mid)) l=mid;else r=mid;
}
cout<<l;return 0;
}

P1419 寻找段落 - 洛谷

 二分答案+优先队列优化动态规划

#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll n,s,t,a[100005];
double b[100005],sum[100005];
deque<int>q;
void put(int x){while(!q.empty()&&sum[q.back()]>=sum[x]){q.pop_back();}q.push_back(x);
}
void out(int x){while(!q.empty()&&q.front()<x){q.pop_front();}
}
bool check(double x){sum[0]=0;for(int i=1;i<=n;i++){b[i]=a[i]-x;sum[i]=sum[i-1]+b[i];}	q.clear();int now=0;for(int i=1;i<=n;i++){while(now<=i-s){put(now++);}out(i-t);if(!q.empty()){if(sum[i]-sum[q.front()]>=0) return true; }}return false;
} 
int main(){
cin>>n>>s>>t;
for(int i=1;i<=n;i++)cin>>a[i];
double l=-1e4-1,r=1e4+1;
while(l+0.0001<r){double mid=(l+r)/2;if(check(mid)) l=mid;else r=mid;
}
cout<<fixed<<setprecision(3)<<l;return 0;
}

P3382 三分 - 洛谷

 可用求导+二分解决

#include<bits/stdc++.h>
#define ll long long
using namespace std;
int n;
double l,r,a[15];
double f(double x){double ans=0,temp=1;for(int i=n+1;i>=1;i--){ans+=a[i]*temp;temp*=x;}return ans;
}
double dao(double x){return 1000000*(f(x+0.000001)-f(x));
}
int main(){
cin>>n>>l>>r;
for(int i=1;i<=n+1;i++){cin>>a[i];
}
while(l+0.000001<r){double mid=(l+r)/2;if(dao(mid)>0) l=mid;else r=mid;
}
cout<<fixed<<setprecision(5)<<l;return 0;
}

P1883 【模板】三分 | 函数 - 洛谷

 直接模板

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;int n, t;
double a[10005], b[10005], c[10005];double f(double x) {double ans = a[1] * x * x + b[1] * x + c[1];for (int i = 2; i <= n; i++) {double val = a[i] * x * x + b[i] * x + c[i];if (val > ans) ans = val;}return ans;
}void work() {cin >> n;for (int i = 1; i <= n; i++) {cin >> a[i] >> b[i] >> c[i];}double l = 0.0, r = 1000.0;while (r - l > 1e-10) {double m1 = l + (r - l) / 3.0;double m2 = r - (r - l) / 3.0;if (f(m1) > f(m2)) l = m1;else r = m2;}double ans = f((l + r) / 2);if (fabs(ans) < 1e-8) ans = 0.0; // 避免输出-0.0000printf("%.4f\n", ans);
}int main() {cin >> t;while (t--) {work();}return 0;
}