http://acm-icpc.aitea.net/index.php?2016%2FPractice%2F%E6%A8%A1%E6%93%AC%E5%9C%B0%E5%8C%BA%E4%BA%88%E9%81%B8%2F%E8%AC%9B%E8%A9%95
C.We don‘t wanna work!
@siludose 你要的代码，做好了参考看
SB模拟，xjb模拟
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <cstring>
#include <queue>
#include <bitset>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
const int MAXN = 1e6+5;
struct Node
{
string s;
int time;
int d;
} a[MAXN];
bool cmp(Node a, Node b)
{
if(a.d == b.d)
return a.time > b.time;
return a.d > b.d;
}
struct cmp2
{
bool operator()(const Node a, const Node b)const
{
if(a.d == b.d)
return a.time > b.time;
return a.d > b.d;
}
};
struct cmp1
{
bool operator()(const Node a, const Node b)const
{
if(a.d == b.d)
return a.time < b.time;
return a.d < b.d;
}
};
set<Node,cmp1>se1;
set<Node,cmp2>se2;
map<string,Node>mp;
string s;
int main()
{
int n, m;
while(~scanf("%d",&n))
{
se1.clear();
se2.clear();
mp.clear();
double tp = 1.0*n*0.2;
int nn = (int)tp;
int tn = n;
for(int i=1; i<=tn; i++)
{
a[i].time = i;
cin>>a[i].s>>a[i].d;
mp[a[i].s] = a[i];
}
sort(a+1, a+1+tn, cmp);
for(int i=1; i<=nn; i++)
se1.insert(a[i]);
for(int i=nn+1; i<=n; i++)
se2.insert(a[i]);
scanf("%d",&m);
Node tmp;
for(int i=tn+1; i<=tn+m; i++)
{
char op;
cin>>op;
if(op == ‘-‘)
{
cin>>s;
tmp = mp[s];
if(se1.erase(tmp)) nn--;
se2.erase(tmp);
if(nn>(int)(1.0*(n-1)*0.2))
{
nn--;
tmp=*se1.begin();
se1.erase(tmp);
se2.insert(tmp);
cout<<tmp.s;
printf(" is not working now.\n");
}
n--;
if(nn<(int)(1.0*(n)*0.2))
{
nn++;
tmp=*se2.begin();
se1.insert(tmp);
se2.erase(tmp);
cout<<tmp.s;
printf(" is working hard now.\n");
}
}
else///++
{
cin>>a[i].s>>a[i].d;
a[i].time=i;
mp[a[i].s]=a[i];
//cout<<nn<<" "<<n<<endl;
if(nn<(int)(1.0*(n+1)*0.2))///+0.2
{
if(a[i].d>(*se2.begin()).d||a[i].d==(*se2.begin()).d&&a[i].time>(*se2.begin()).time)
{
se1.insert(a[i]);
cout<<a[i].s;
printf(" is working hard now.\n");
}
else
{
tmp=*se2.begin();
se2.erase(tmp);
se1.insert(tmp);
se2.insert(a[i]);
cout<<a[i].s;
printf(" is not working now.\n");
cout<<tmp.s;
printf(" is working hard now.\n");
}
nn++;
//cout<<"nn"<<nn<<endl;
}
else///=0.2
{
if(nn!=0)
{
tmp=*se1.begin();
if(a[i].d>tmp.d||a[i].d==tmp.d&&a[i].time>tmp.time)
{
se1.erase(tmp);
se1.insert(a[i]);
se2.insert(tmp);
cout<<a[i].s;
printf(" is working hard now.\n");
cout<<tmp.s;
printf(" is not working now.\n");
}
else
{
se2.insert(a[i]);
// se2.erase(tmp);
//se1.insert(tmp);
cout<<a[i].s;
printf(" is not working now.\n");
//cout<<tmp.s;
//printf(" is working hard now.\n");
///
}
}
else
{
tmp=*se2.begin();
if((int)(1.0*(n+1)*0.2)>0)
{
if(a[i].d>tmp.d||a[i].d==tmp.d&&a[i].time>tmp.time)
{
se1.insert(a[i]);
cout<<a[i].s;
printf(" is working hard now.\n");
}
else
{
se2.erase(tmp);
se2.insert(a[i]);
se1.insert(tmp);
cout<<a[i].s;
printf(" is not working now.\n");
cout<<tmp.s;
printf(" is working hard now.\n");
}
}
else
{
se2.insert(a[i]);
cout<<a[i].s;
printf(" is not working now.\n");
}
}
}
n++;
}///++
}
}
return 0;
}
C
E.Similarity of Subtrees
真心佩服那帮用递归栈都能不爆栈的大神。。。不说了，伤心题
题目要求求出相似点对对数，相似指2对子树在不同且对应深度的结点都一样
可以看做可加的向量：(d0,d1, ... ,dn)其中dk是指深度为k的结点个数
父结点的向量是其本身(1,0,0, ... )+(0,所有子结点的向量之和)
不过要是直接向量上会o(n^2)滚粗
所以搞成hash值:d0*a^0+d1*a^1+d2*a^2+...
然后发现数字太大了，要模
a>100000就可以，没有哪个向量分量超过100000
对结果模的数m>1000000000，且必须是素数，感觉不设这么大会有hash冲突
可以map搞o(nlogn)或者hash_map搞o(n)
比赛中要注意要是这种数据都能爆栈，就用非递归+stack<int>硬杠
此题还可以bfs硬撑
#include <iostream>
#include <stdio.h>
#include <queue>
#include <string.h>
#include <vector>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const LL maxn=1e6+10,p=9901,mod=1e9+7;
vector<LL>G[maxn];
LL hash_v[maxn];
map<LL,LL>mp;
map<LL,LL>::iterator it;
void dfs(LL u)
{
hash_v[u]=1;
for(LL i=0; i<G[u].size(); i++)
{
LL v=G[u][i];
dfs(v);
hash_v[u]=(hash_v[u]+hash_v[v]*p)%mod;
}
mp[hash_v[u]]++;
}
int main()
{
///freopen("in.txt","r",stdin);
LL n;
LL u,v;
while(scanf("%lld",&n)!=-1)
{
for(LL i=0; i<=n; i++)
G[i].clear();
mp.clear();
for(LL i=1; i<n; i++)
{
scanf("%lld%lld",&u,&v);
G[u].push_back(v);
}
dfs(1);
LL ans=0;
for(it=mp.begin(); it!=mp.end(); it++)
ans+=it->second*(it->second-1)/2;
printf("%lld\n",ans);
}
return 0;
}
F.Escape from the Hell
题意：有人在悬崖上的1条绳子上，
有n罐能量饮料，喝的当天可以向上爬ai米，但是若没有爬到顶点，之后会下滑bi米，不喝则不会动
与此同时，有只蜘蛛第i天会不下滑地向上ci米，且上升到人身上就会致死
问人能否逃离，第几天逃离
题解：枚举最后1天喝哪种饮料，并从饮料集合去除
把其他饮料按照a(i)-b(i)的大小排序，负数的一定没用
设sdis(i)=sdis(i-1)+c(i) pdis(i)=pdis(i)+a(i)-b(i)
找到最小的x使得
任何i<x都是pdis(i)>sdis(i)并且pdis(x-1)+a(x)>=L，就是逃离成功
枚举过的那个元素不用再重复枚举
G.Shere the Ruins Presevation
题意：把点集以1条平行于y轴且不相切任何点的直线分成2个点集，然后用栅栏把2个点集围成最小面积
题解：感觉就是对整个点集以凸包的围法画边，把整条x轴按照点集在x上的分量离散化
把整个图形按顺序分割出三角形，再把那些三角形根据占用x轴的区间把面积写在树状数组
查询时相邻区间查询因为分割而少掉的面积，最大的那个减去总面积就是解
andrew‘s monotone chain?