题意
分析
考场做法
考虑二分答案,R开到1e9就能过了。
判断答案合法,就判断时间和是否超过拥有的时间就行了。但要把di从小到大排序,不然容易验证贪心是错的。
时间复杂度\(O(n \log n)\)
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#include<string>
#include<vector>
#include<list>
#include<deque>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<complex>
#include<cassert>
#define rg register
#define il inline
#define co const
#pragma GCC optimize ("O0")
using namespace std;
template<class T> il T read()
{
T data=0;
int w=1;
char ch=getchar();
while(!isdigit(ch))
{
if(ch=='-')
w=-1;
ch=getchar();
}
while(isdigit(ch))
data=10*data+ch-'0',ch=getchar();
return data*w;
}
template<class T> il T read(T&x)
{
return x=read<T>();
}
typedef long long ll;
const int INF=0x7fffffff;
const int MAXN=1e5+7;
int n;
struct $
{
int t,d;
bool operator<(const $&rhs)const
{
return d<rhs.d;
}
}a[MAXN];
bool judge(int M)
{
ll sum=M;
for(int i=1;i<=n;++i)
{
sum+=a[i].d-a[i-1].d;
sum-=a[i].t;
if(sum<0)
return 0;
}
return 1;
}
int main()
{
freopen("ming.in","r",stdin);
freopen("ming.out","w",stdout);
read(n);
for(int i=1;i<=n;++i)
{
read(a[i].t);read(a[i].d);
}
sort(a+1,a+n+1);
int L=0,R=1e9,ans;
while(L<=R)
{
int M=(L+R)>>1;
if(judge(M))
ans=M,R=M-1;
else
L=M+1;
}
printf("%d\n",ans);
// fclose(stdin);
// fclose(stdout);
return 0;
}
标解
旁边的大佬L君:类似B君讲过的一道题,发现可分析贡献排序,时间复杂度\(O(n \log n)\)
#include<bits/stdc++.h>
using namespace std;
#define gc c=getchar()
#define r(x) read(x)
#define ll long long
template<typename T>
inline void read(T&x){
x=0;T k=1;char gc;
while(!isdigit(c)){if(c=='-')k=-1;gc;}
while(isdigit(c)){x=x*10+c-'0';gc;}x*=k;
}
const int N=1e5+7;
struct Data{
ll t,d;
}A[N];
inline bool operator < (const Data &a,const Data &b){
return a.d<b.d;
}
int main(){
freopen("ming.in","r",stdin);
freopen("ming.out","w",stdout);
int n;r(n);
for(int i=0;i<n;++i)r(A[i].t),r(A[i].d);
sort(A,A+n);
ll ans=0,tim=0;
for(int i=0;i<n;++i){
tim+=A[i].t;
ans=max(ans,tim-A[i].d);
}
printf("%lld\n",ans);
}