The 2014 ACMICPC Asia Regional Beijing Online
【C】-_-///不懂
【D】数论,概率密度
【H】区间维护+线段树+DFS序(可以看看)
【I】BFS地图题(当时好多人坑在摄像头上面了,现在有一点点思路,分层图,一会看看)
【J】-_-/////
貌似除了这两题巨坑的,剩下的都有能出的可能性
【A】
Always Cook Mushroom
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
【Sample Output】
Case #: Case #:
【题意】 给出一张最大1000*1000的图,给出一些询问,每次询问给出一个斜率和x,要求三角内的所有点的和。
【分析】
题目意思比较裸,直观地想到这道题目的难度肯定不是在理解上,应该是数据比较大太裸的算法不可能卡过。
果然试了各种,唯一想到的树状数组也加上去了,还是一直TLE没能在比赛过程中出解。
赛后整理终于明白了,这道题目的关键在于如何把一张二维的图转化为一维来解决,本题给我的启发是:累加的题目如果要避免反复加引起的重复的话,树状数组是必须的,时间非常优秀,然而树状数组毕竟是一维的,必须想到办法解决二维压成一维之后的后效性等等问题才能发挥出最大的效果。
最后的算法:
二维转一维:将最大1000*1000个点按照斜率排序。对于每次的询问也是按照斜率排序。
想象有一根轴,从x轴的0°角开始逆时针扫整个平面,扫到的点按照x坐标加入树状数组,直到扫到与一条询问线重合,则对该次询问求sum(x)。这样就能避免了重复操作。把结果按照原始顺序重新输出即可。
/* ***********************************************
MYID : Chen Fan
LANG : G++
PROG : HDU5032
************************************************ */ #include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm> using namespace std; long long c[];
int outp[];
long long ou[]; typedef struct poin
{
int x,y;
long long v;
double ang;
} point; point poi[];
int totp; typedef struct nod
{
int a,b,x,src;
double ang;
} node; node lis[]; bool op1(point a,point b)
{
if (a.ang==b.ang) return a.x<b.x;
else return a.ang<b.ang;
} bool op2(node a,node b)
{
if (a.ang==b.ang) return a.a<b.a;
else return a.ang<b.ang;
} int lowbit(int s)
{
return s&-s;
} void update(int s,long long x)
{
while (s<=)
{
c[s]+=x;
s+=lowbit(s);
}
} long long sum(int s)
{
long long t=;
while (s>)
{
t+=c[s];
s-=lowbit(s);
}
return t;
} int main()
{
freopen("test.txt","r",stdin); int t; totp=;
for (int i=;i<=;i++)
for (int j=;j<=;j++)
{
totp++;
poi[totp].x=i;
poi[totp].y=j;
poi[totp].ang=(double)j/i;
}
sort(&poi[],&poi[+totp],op1); scanf("%d",&t);
for (int tt=;tt<=t;tt++)
{
int A,B;
scanf("%d%d",&A,&B);
memset(c,,sizeof(c)); for (int i=;i<=totp;i++) poi[i].v=(poi[i].x+A)*(poi[i].y+B); printf("Case #%d:\n",tt); int m;
scanf("%d",&m);
for (int i=;i<=m;i++)
{
scanf("%d%d%d",&lis[i].a,&lis[i].b,&lis[i].x);
lis[i].ang=(double)lis[i].b/lis[i].a;
lis[i].src=i;
}
sort(&lis[],&lis[+m],op2);
for (int i=;i<=m;i++) outp[lis[i].src]=i; for (int i=,j=;i<=m;i++)
{
while (j<=totp&&lis[i].ang-poi[j].ang>=)
{
update(poi[j].x,poi[j].v);
j++;
}
ou[i]=sum(lis[i].x);
} for (int i=;i<=m;i++) printf("%lld\n",ou[outp[i]]);
} return ;
}
【启发】
既然已经想到了树状数组,就应该多想想树状数组的性质,本题是利用了极角的特性二维化一维,思维上非常地巧妙。
树状数组这一块另外再补补,尤其是多维的情况
【B】
Building
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Special Judge
【Sample Output】
Case #:
101.3099324740
Case #:
90.0000000000
Case #:
78.6900675260
【说明】
不太擅长几何题,好在队友给力能解决,日后慢慢看......0.0,嘿嘿
【E】
Explosion
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others) Total
In the first line of each test case, there is an integer N (N<=1000) indicating the number of rooms.
The following N lines corresponde to the rooms from 1 to N. Each line begins with an integer k (0<=k<=N) indicating the number of keys behind the door. Then k integers follow corresponding to the rooms these keys can open.
【Sample Output】
Case #: 1.00000
Case #: 3.00000
【题意】
/* ***********************************************
MYID : Chen Fan
LANG : G++
PROG : HDU5036
************************************************ */ #include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <bitset> using namespace std; bitset<> a[]; int main()
{
int t;
scanf("%d",&t);
for (int tt=;tt<=t;tt++)
{
int n;
scanf("%d",&n);
for (int i=;i<=n;i++)
{
a[i].reset();
a[i][i]=true;
} for (int i=;i<=n;i++)
{
int m;
scanf("%d",&m);
for (int j=;j<=m;j++)
{
int k;
scanf("%d",&k);
a[i][k]=true;
}
} for (int i=;i<=n;i++)
for (int j=;j<=n;j++)
if (a[j][i]) a[j]|=a[i]; double ans=;
for (int i=;i<=n;i++)
{
int tot=;
for (int j=;j<=n;j++)
if (a[j][i]) tot++;
ans+=1.0/tot;
} printf("Case #%d: %.5lf\n",tt,ans);
} return ;
}
有关bitset的用法,详见另外整理的一篇博文:
http://www.cnblogs.com/jcf94/p/3997908.html
【F】
Frog
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
The river could be considered as an axis.Matt is standing on the left bank now (at position 0). He wants to cross the river, reach the right bank (at position M). But Matt could only jump for at most L units, for example from 0 to L.
As the God of Nature, you must save this poor frog.There are N rocks lying in the river initially. The size of the rock is negligible. So it can be indicated by a point in the axis. Matt can jump to or from a rock as well as the bank.
You don't want to make the things that easy. So you will put some new rocks into the river such that Matt could jump over the river in maximal steps.And you don't care the number of rocks you add since you are the God.
Note that Matt is so clever that he always choose the optimal way after you put down all the rocks.
For each test case, the first line contains N, M, L (0<=N<=2*10^5,1<=M<=10^9, 1<=L<=10^9).
And in the following N lines, each line contains one integer within (0, M) indicating the position of rock.
【Sample Output】
Case #:
Case #:
【题意】
【分析】
【G】
Grade
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
The first line of each test cases contains one integers N (1<=N<=10^6),denoting the number of the mushroom.
The second line contains N integers, denoting the weight of each mushroom. The weight is greater than 0, and less than 200.
The first line contains "Case #x:", where x is the case number (starting from 1)
The second line contains the mode of the grade of the given mushrooms. If there exists multiple modes, output them in ascending order. If there exists no mode, output “Bad Mushroom”.
【Sample Output】
Case #: Case #:
Bad Mushroom
Case #:
【分析】
排序+判断,有些细节上的小问题要多注意一下
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