【CF1257D】Yet Another Monster Killing Problem【贪心】
题意:给定一些怪物,每天可以选一个勇士进去打怪,每个勇士每天只能打不超过si个怪物,每个勇士只能打能力值≤pi的怪物,问最少多少天打完所有怪物
题解:贪心,每天尽可能多的去打怪,那么存一个对于长度为i的怪物可以用的最大的能力值是多少,每天枚举长度,尽可能长的去打怪即可
代码:
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#define ll long long
using namespace std;
int T,n,m,ans;
int a[],mx[];
struct node
{
int p,s;
}b[];
bool cmp(const node &T1,const node &T2){return T1.s>T2.s;}
int main()
{
scanf("%d",&T);
while(T--)
{
for(int i=;i<=n;i++)mx[i]=;
scanf("%d",&n);
for(int i=;i<=n;i++)scanf("%d",&a[i]);
scanf("%d",&m);
for(int i=;i<=m;i++)scanf("%d%d",&b[i].p,&b[i].s);
sort(b+,b++m,cmp);
int j=,t=;
for(int i=n;i>;i--)
{
while(j<=m && b[j].s>=i)
{
t=max(t,b[j].p);
j++;
}
mx[i]=t;
}
ans=;j=;t=;
for(int i=;i<=n;i++)
{
t=max(t,a[i]);
if(t>mx[j])
{
//printf("%d %d %d %d\n",i,j,a[i],mx[j]);
if(j==){ans=-;break;}
ans++;j=;i--;t=;
}
else j++;
}
if(ans>-)ans++;
//for(int i=1;i<=n;i++)printf("%d ",mx[i]);
printf("%d\n",ans);
}
return ;
}
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