T2691 桶哥的问题——送桶

这个题其实不难,就是按照结束时候的顺序从大到小走一遍,能送的就送,如果区间不重合就更新一下

代码:

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<time.h>
using namespace std;
typedef long long ll;
ll read(){
    int ans=0;
    char last=' ',ch=getchar();
    while(ch<'0' || ch>'9')last=ch,ch=getchar();
    while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();
    if(last=='-')ans=-ans;
    return ans;
}

const int N=1e6+5;

//ll a[N],b[N],c[N],n,m,ans;
//ll sum;

ll t,n;

struct qwq
{
    ll a,b;
}num[N];

bool cmp(qwq x,qwq y)
{
    return x.b>y.b;
}

int main()
{
    
    n=read();
    for(int i=1;i<=n;i++)
    {
        num[i].a=read();
        num[i].b=read();
        
    }
    sort(num+1,num+1+n,cmp);
    t=num[1].b;
    for(int i=1;i<=n;i++)
    {
        if(t>num[i].b)t=num[i].b;
        t-=num[i].a;
    }
    
    cout<<t;
}
原文地址:https://www.cnblogs.com/lcezych/p/10939838.html