洛谷P2949 工作调度Work Scheduling [USACO09OPEN] 贪心

时间:2023-03-09 07:55:00
洛谷P2949 工作调度Work Scheduling [USACO09OPEN] 贪心

正解:贪心+并查集(umm不用并查集也成qwq

解题报告:

水题?主要感觉想到了俩方法然后还只实现了一个,怕忘了所以想着开个新坑记录下qwq

然后先放下传送门QAQ(哦这题和supermarket,双倍经验qwq

第一个是贪心+单调队列,和蔬菜&supermarket这两题都挺像的一个想法,就是能往后安排往后安排,因此就先按时间排序然后用个priority_queue就过去辽

第二个是也是贪心但是也许复杂度好看些?不清楚诶QAQ就是用个并查集合并日期,就有点像那个,疯狂的馒头?好了没了,over,有时间再贴代码QAQ

算了先放下第一个的代码趴qwq第二个还没打等打了再放上来QAQ

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rp(i,x,y) for(register ll i=x;i<=y;++i) const ll N=+;
ll n,ans;
struct node{ll p,d;}wk[N];
priority_queue<node>Q;
bool operator < (node gd,node gs){return gd.p>gs.p;} ll read()
{
char ch=getchar();ll x=;bool y=;
while(ch!='-' && (ch>'' || ch<''))ch=getchar();
if(ch=='-')ch=getchar(),y=;
while(ch>='' && ch<='')x=(x<<)+(x<<)+(ch^''),ch=getchar();
return y?x:-x;
} inline bool cmp(node gd,node gs){return gd.d<gs.d;} int main()
{
n=read();rp(i,,n)wk[i].d=read(),wk[i].p=read();sort(wk+,wk++n,cmp);
rp(i,,n){if(Q.size()>=wk[i].d){int t=Q.top().p;if(t<wk[i].p)ans+=wk[i].p-t,Q.pop(),Q.push(wk[i]);}else ans+=wk[i].p,Q.push(wk[i]);}
printf("%lld\n",ans);
return ;
}

然后关于方法二,,,,它只能过三个点,一个点RE其他点WA,,,然后supermarket的话从数据范围来说应该不会WA?然而我用的小号不能提交,,,QAQ

然后我就交到vjudge上

然后WA了

然后udebug上居然没有数据下

就很生气

就不想做了 咕咕咕

算了先放下代码存着8

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rp(i,x,y) for(register ll i=x;i<=y;++i) const ll N=+;
ll n,ans,fa[N],mx;
struct node{ll p,d;}wk[N]; ll read()
{
char ch=getchar();ll x=;bool y=;
while(ch!='-' && (ch>'' || ch<''))ch=getchar();
if(ch=='-')ch=getchar(),y=;
while(ch>='' && ch<='')x=(x<<)+(x<<)+(ch^''),ch=getchar();
return y?x:-x;
} inline bool cmp(node gd,node gs){return gd.p>gs.p;}
inline ll fd(ll x){return fa[x]==x?fa[x]:fa[x]=fd(fa[x]);} int main()
{
while(cin>>n)
{
ans=;
rp(i,,n)wk[i].p=read(),wk[i].d=read(),mx=max(mx,wk[i].d);sort(wk+,wk++n,cmp);rp(i,,mx)fa[i]=i;
rp(i,,n){ll t=fd(wk[i].d);if(t==)continue;fa[wk[i].d]=wk[i].d-;ans+=wk[i].p;}
printf("%lld\n",ans);
}
return ;
}

QAQQQ