CF720A Closing ceremony 贪心

正解:贪心

解题报告:

传送门!

先考虑如果只有一列怎么搞?那就肯定是尽量走到最远的地方

然后用点儿类似的思想,现在考虑有两列的情况QAQ

为了方便表述,这里给每个位置两个值,a表示离一号入口的距离,b表示离二号入口的距离

可以先把从一号入口进去的安排了,所以考虑按体力从小往大考虑,然后在能满足a的情况下尽量选b比较大的位置

然后再处理二号入口的,就按体力从大往小考虑一下就好

具体实现的话就先把所有坐标按a排序,然后满足的就加入堆,堆的内部按b排序就好

over

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define gc getchar()
#define ri register int
#define rc register char
#define rb register bool
#define rp(i,x,y) for(ri i=x;i<=y;++i)
#define my(i,x,y) for(ri i=x;i>=y;--i)

const int N=+;
int n,m,k,a[N],nw=,nod_cnt;
struct node{int x,y,disa,disb;}nod[N];
il bool operator < (node gd,node gs){return gd.disb<gs.disb;}
priority_queue<node>Q;

il int read()
{
    rc ch=gc;ri x=;rb y=;
    while(ch!='-' && (ch>'' || ch<''))ch=gc;
    if(ch=='-')ch=gc,y=;
    while(ch>='' && ch<='')x=(x<<)+(x<<)+(ch^''),ch=gc;
    return y?x:-x;
}
il bool cmp(node gd,node gs){return gd.disa<gs.disa;}
il bool cmq(ri gd,ri gs){return gd>gs;}

int main()
{
//     freopen("720a.in","r",stdin);freopen("720a.out","w",stdout);
    n=read();m=read();rp(i,,n)rp(j,,m)nod[++nod_cnt]=(node){i,j,i+j,i+m+-j};sort(nod+,nod++nod_cnt,cmp);
    k=read();rp(i,,k)a[i]=read();sort(a+,a++k);
    rp(i,,k){while(a[i]>=nod[nw].disa && nw<=nod_cnt)Q.push(nod[nw++]);if(Q.empty())return printf("NO\n"),;Q.pop();}
    while(nw<=nod_cnt)Q.push(nod[nw++]);
    k=read();rp(i,,k)a[i]=read();sort(a+,a++k);
//    printf("nmdwsm??? %d\n",Q.size());
//    while(!Q.empty())printf("%d\n",Q.top().disb),Q.pop();
    my(i,k,){if(a[i]<Q.top().disb)return printf("NO\n"),;Q.pop();}//printf("dis=%d\n",Q.top().disb);}
    printf("YES\n");
    return ;
}

然后这儿是代码QwQ!