2016-2017 ACM-ICPC, Asia Tsukuba Regional Contest
题目链接：http://codeforces.com/gym/101158
A
签到题
#include
using namespace std;
int a[200010],b[200010];
int main()
{
int n,m;
scanf("%d%d",&n,&m);
memset(a,0,sizeof(a));
for(int i = 1;i <= m;i++)
scanf("%d",&b[i]);
for(int i = m;i >= 1;i--)
if(a[b[i]]==0){
a[b[i]] = 1;
printf("%d\n",b[i]);
}
for(int i = 1;i <= n;i++)
if(a[i]==0)
printf("%d\n",i);
} 模拟一下，暴一暴就过了啊？
#include
using namespace std;
int sv[12][12];
int Xor(int i,int j)
{
return sv[i][j];
}
int get(string tmp)
{
int la=0;
for(int i=0;i<4;i++)
{
int id=tmp[i]-'0';
la=Xor(la,id);
}
return la;
}
int ans=0;
bool judge(int now)
{
string tmp;
for(int i=0;i<4;i++)
{
tmp+=now%10+'0';
now/=10;
}
reverse(tmp.begin(),tmp.end());
tmp+=get(tmp)+'0';
for(int i=0;i<5;i++)
{
int id=tmp[i]-'0';
for(int j=0;j<10;j++)
{
string S=tmp;
if(j==id)
{
continue;
}
S[i]=(j+'0');
int la=get(S);
if(!Xor(la,S[4]-'0'))
return true;
}
}
for(int i=0;i<4;i++)
{
string S=tmp;
swap(S[i],S[i+1]);
int la=get(S);
if(S==tmp)
continue;
if(!Xor(la,S[4]-'0'))
return true;
}
return false;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
for(int i=0;i<10;i++)
{
for(int j=0;j<10;j++)
{
scanf("%d",&sv[i][j]);
}
}
//judge(2016);
for(int i=0;i<10000;i++)
{
ans+=judge(i);
}
printf("%d\n",ans);
return 0;
} 继续模拟
#include
using namespace std;
const int MAXN=2e5+5;
struct node{
int x,y;
}p[MAXN];
struct edge
{
int x,ca;
edge(){}
edge(int _x,int _ca):x(_x),ca(_ca){}
bool operator < (const edge &a)const
{
return x DO[MAXN],UP[MAXN];
int cl[MAXN],cs[MAXN];
int main(){
int n,m;
scanf("%d%d",&n,&m);
for(int i = 0;i < m;i++)
{
scanf("%d%d",&p[i].x,&p[i].y);
DO[p[i].y].push_back(edge(p[i].x,0));
UP[p[i].y+1].push_back(edge(p[i].x,0));
}
DO[0].push_back(edge(0,0));
UP[n+1].push_back(edge(0,0));
for(int i=1;i<=n;i++)
{
cs[i]=cl[i]=1;
DO[i].push_back(edge(0,0));
UP[i].push_back(edge(0,0));
sort(DO[i].begin(),DO[i].end());
sort(UP[i].begin(),UP[i].end());
}
int at=0;
for(int i=1;i<=n;i++)
{
for(int j=1;j :: iterator it=upper_bound(DO[i-1].begin(),DO[i-1].end(),edge(at,0));
it--;
DO[i][j].ca=it->ca+1;
cl[i]=max(DO[i][j].ca,cl[i]);
}
at=1e9;
vector :: iterator it=upper_bound(DO[i-1].begin(),DO[i-1].end(),edge(at,0));
it--;
cl[i]=max(it->ca+1,cl[i]);
}
for(int i=n;i>=1;i--)
{
for(int j=1;j :: iterator it=upper_bound(UP[i+1].begin(),UP[i+1].end(),edge(at,0));
it--;
UP[i][j].ca=it->ca+1;
cs[i]=max(UP[i][j].ca,cs[i]);
}
at=1e9;
vector :: iterator it=upper_bound(UP[i+1].begin(),UP[i+1].end(),edge(at,0));
it--;
cs[i]=max(it->ca+1,cs[i]);
}
for(int i=1;i<=n;i++)
{
//printf("%d %d ",cl[i],cs[i]);
printf("%d ",cl[i]+cs[i]-1);
}
} 花式暴力？
#include
#include
#include
#include
#include
#include
#include python暴力？
s = raw_input()
ans = {}
op = ['+','-','*','==','(',')','0','1']
ops = ['0','0','0','0','0','0','0','0']
inuse = [0,0,0,0,0,0,0,0]
cnt=0
def check():
global ops
global cnt
str = ''
be = 0;
cnks = 0
dict = {}
for chr in s:
if not chr.isalpha():
if chr=='=':
str += chr
str += chr
continue
if not dict.has_key(chr):
if be == 8:
return;
dict[chr]=be
be+=1
str+= ops[dict[chr]]
tar = 0
k= len(str)
lays = 0
for i in str:
if i == '(':
lays+=1
if i == ')':
lays -=1
if i =='=':
if lays != 0:
return
for i in range(len(str)):
if str[i]=='(' and i != k-1 and str[i+1]==')' :
return
if str[i]=='*' and i != k-1 and str[i+1]=='*':
return
if str[i]=='+':
if i == 0 or (str[i-1]!='0' and str[i-1]!='1' and str[i-1]!=')'):
return
if str[i]=='=':
tar+=1
if i!=k-1 and str[i]=='0' and (i==0 or (str[i-1] != '0' and str[i-1]!= '1')) and (str[i+1]=='0' or str[i+1]=='1'):
return
if tar!= 2:
return
execs = ''
tar = 1
for i in str:
if (i == '0' or i == '1'):
if tar==1:
execs += '0b'
execs += i
tar=0
else:
execs += i
else:
execs += i
tar=1
try:
if eval(execs)==True:
if not ans.has_key(str):
ans[str]=1
except:
cnt = cnt
def dfs(lay):
global ops
global op
global inuse
if lay == 8:
check()
return;
else:
for i in range(8):
if inuse[i]==0:
ops[lay]=op[i]
inuse[i]=1
dfs(lay+1)
inuse[i]=0
return;
dfs(0)
print len(ans)bitset优化暴力
#include
#define xx first
#define yy second
#define mp make_pair
using namespace std;
const int MAXN=1005;
vector sv[MAXN][MAXN];
vector > edge[MAXN];
bitset G[MAXN];
map Hash;
int tot=0;
bool vis[MAXN];
int getid(string now)
{
int &id=Hash[now];
if(id==0)
id=++tot;
return id;
}
void dfs(int a,int b)
{
G[a][b]=1;
for(auto it:sv[a][b])
{
if(vis[it])
continue;
vis[it]=true;
for(auto itt:edge[it])
{
int p,q;
p=itt.xx,q=itt.yy;
if(!G[p][q])
dfs(p,q);
}
}
for(int i=1;i<=tot;i++)
{
if(G[i][a])
{
bitset bs=((~G[i])&G[b]);
G[i]|=G[b];
for(int j=1;j<=tot;j++)
{
if(bs[j])
{
for(auto it:sv[i][j])
{
if(vis[it])
continue;
vis[it]=true;
for(auto itt:edge[it])
{
int p,q;
p=itt.xx,q=itt.yy;
if(!G[p][q])
dfs(p,q);
}
}
}
}
}
}
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n;
string st,en;
cin>>st>>en;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
int k;
scanf("%d",&k);
for(int j=1;j<=k;j++)
{
string a,b;
cin>>a>>b;
int ida=getid(a),idb=getid(b);
sv[ida][idb].push_back(i);
edge[i].push_back(mp(ida,idb));
}
}
int start=getid(st),end=getid(en);
for(int i=1;i<=tot;i++)
G[i][i]=1;
dfs(start,end);
for(int i=1;i<=tot;i++)
{
for(int j=i+1;j<=tot;j++)
{
if(G[i][j]&&G[j][i])
{
return 0*printf("No\n");
}
}
}
printf("Yes\n");
return 0;
} G
思路很巧妙的数学题。。。（具体看题解），set暴力。。。（从cls那里学到了先进的set用法？）
#include
using namespace std;
set sv;
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n;
int now=1;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
int x;
scanf("%d",&x);
if(xx)
puts("No");
else
{
while(!sv.insert(x).second)
{
sv.erase(x);
x--;
}
while(sv.count(now))
now++;
puts("Yes");
}
}
return 0;
} 很有意思图论题目，可以转换为一个DAG图套树形DP，膜拜了wode cude dalao的代码才看懂。。。
#include
#define xx first
#define yy second
#define mp make_pair
using namespace std;
const int MAXN=1e5+5;
vector > sv[3];
vector e[MAXN],E[MAXN],et[MAXN];
//Q
int fa[MAXN],in[MAXN],dp[MAXN],U[MAXN],D[MAXN],A[MAXN];
void Init(int n)
{
for(int i=1;i<=n;i++)fa[i]=i;
}
int Find(int x)
{
return x==fa[x] ? x : fa[x]=Find(fa[x]);
}
bool Union(int x,int y)
{
x=Find(x),y=Find(y);
if(x==y)return 0;
return fa[x]=y,1;
}
struct node
{
multiset s;
int get()
{
return *s.rbegin();
}
node(){}
node(int val)
{
s.insert(val);
}
void operator += (const int &a)
{
s.insert(a);
}
int operator - (const int &a)
{
int ret;
auto it=s.find(a);
s.erase(it);
ret=get();
s.insert(a);
return ret;
}
};
node tree[MAXN];
void dfsD(int now,int f)
{
tree[now]=node(dp[now]);
for(auto i:e[now])
{
if(i==f)
continue;
dfsD(i,now);
tree[now]+=(D[i]+1);
}
D[now]=tree[now].get();
}
void dfsU(int now,int f)
{
if(f)
tree[now]+=(U[now]+1);
A[now]=tree[now].get();
for(auto i:e[now])
{
if(i==f)
continue;
U[i]=tree[now]-(D[i]+1);
dfsU(i,now);
}
}
void dfsT(int now,int f)
{
dp[now]=A[now];
for(auto i:et[now])
{
dp[i]=max(dp[i],dp[now]+1);
}
for(auto i:e[now])
{
if(i==f)
continue;
dfsT(i,now);
}
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n,m;
scanf("%d%d",&n,&m);
Init(n);
for(int i=1;i<=m;i++)
{
int u,v,op;
scanf("%d%d%d",&u,&v,&op);
sv[op].push_back({u,v});
if(op==2)
{
if(!Union(u,v))
{
return 0*puts("Infinite\n");
}
e[u].push_back(v);
e[v].push_back(u);
}
}
for(auto i:sv[1])
{
int u,v;
u=i.xx,v=i.yy;
u=Find(u),v=Find(v);
E[u].push_back(v);
in[v]++;
}
vector Q;
int tot=0;
for(int i=1;i<=n;i++)
{
if(i==Find(i))
{
tot++;
if(in[i]==0)
Q.push_back(i);
}
}
for(int i=0;i 这竟然，是个数学题啊？
#include
using namespace std;
int gcd(int a,int b,int &x1,int &y1){
if (b==0){
x1=1,y1=0;
return a;
}
int q=gcd(b,a%b,y1,x1);
y1-=a/b*x1;
return q;
}
int main(){
int n;
int x,y,x1,y1;
scanf("%d",&n);
while(n--){
scanf("%d%d",&x,&y);
if(x<=50000||y<=50000){
printf("3\n0 0\n%d %d\n",x,y);
if(x0&&y1>0){
x1 -= yy;
y1 -= xx;
}
printf("%d %d\n",abs(y1),abs(x1));
}
}
} 模拟退火暴一暴
#include
#include
#include
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)<(y)?(x):(y))
#define PI acos(-1.0)
#define EPS 1e-10
#include
#include
using namespace std;
inline int DB(double x) {
if(x<-EPS) return -1;
if(x>EPS) return 1;
return 0;
}
struct point {
double x,y;
point() {}
point(double _x,double _y):x(_x),y(_y) {}
point operator-(point a) {
return point(x-a.x,y-a.y);
}
point operator+(point a) {
return point(x+a.x,y+a.y);
}
double operator*(point a) {
return x*a.y-y*a.x;
}
point oppose() {
return point(-x,-y);
}
double length() {
return sqrt(x*x+y*y);
}
point adjust(double L) {
L/=length();
return point(x*L,y*L);
}
point vertical() {
return point(-y,x);
}
double operator^(point a) {
return x*a.x+y*a.y;
}
};
struct segment {
point a,b;
segment() {}
segment(point _a,point _b):a(_a),b(_b) {}
point intersect(segment s) {
double s1=(s.a-a)*(s.b-a);
double s2=(s.b-b)*(s.a-b);
double t=s1+s2;
s1/=t;
s2/=t;
return point(a.x*s2+b.x*s1,a.y*s2+b.y*s1);
}
point vertical(point p) {
point t=(b-a).vertical();
return intersect(segment(p,p+t));
}
int isonsegment(point p) {
return DB(min(a.x,b.x)-p.x)<=0&&
DB(max(a.x,b.x)-p.x)>=0&&
DB(min(a.y,b.y)-p.y)<=0&&
DB(max(a.y,b.y)-p.y)>=0;
}
};
struct circle {
point p;
double R;
};
circle C;
point p[105];
int n;
double cross_area(point a,point b,circle C) {
point p=C.p;
double R=C.R;
int sgn=DB((b-p)*(a-p));
double La=(a-p).length(),Lb=(b-p).length();
int ra=DB(La-R),rb=DB(Lb-R);
double ang=acos(((b-p)^(a-p))/(La*Lb));
segment t(a,b);
point s;
point h,u,temp;
double ans,L,d,ang1;
if(!DB(La)||!DB(Lb)||!sgn) ans=0;
else if(ra<=0&&rb<=0) ans=fabs((b-p)*(a-p))/2;
else if(ra>=0&&rb>=0) {
h=t.vertical(p);
L=(h-p).length();
if(!t.isonsegment(h)||DB(L-R)>=0) ans=R*R*(ang/2);
else {
ans=R*R*(ang/2);
ang1=acos(L/R);
ans-=R*R*ang1;
ans+=R*sin(ang1)*L;
}
} else {
h=t.vertical(p);
L=(h-p).length();
s=b-a;
d=sqrt(R*R-L*L);
s=s.adjust(d);
if(t.isonsegment(h+s)) u=h+s;
else u=h+s.oppose();
if(ra==1) temp=a,a=b,b=temp;
ans=fabs((a-p)*(u-p))/2;
ang1=acos(((u-p)^(b-p))/((u-p).length()*(b-p).length()));
ans+=R*R*(ang1/2);
}
return ans*sgn;
}
double cal_cross(circle C,point p[],int n) {
double ans=0;
int i;
p[n]=p[0];
for(i=0; i 1e-6; i++) {
flag = -1;
for (int k = 0; k < 40; k++) {
double ang = (rand()%1000+1)/1000.0*10*pi;
tmp.x = ans.x + step * cos(ang);
tmp.y = ans.y + step * sin(ang);
C.p = tmp;
if (ncost < cal_cross(C,p,n)) {
ncost = cal_cross(C,p,n);
flag = 1;
ans = tmp;
}
}
if (flag == -1) step *= 0.8;
}
if(ncost>acost){
acost = ncost;
cnt = ans;
}
}
//printf("(%lf,%lf).\n", cnt.x, cnt.y);
printf("%lf\n", acost);
return 0;
} 奇奇怪怪的博弈。。。
#define _CRT_SECURE_NO_WARNINGS
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