test20181024 ming

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);
}