2016-2017 ACM-ICPC, Asia Tsukuba Regional Contest

题目链接：http://codeforces.com/gym/101158

A

签到题

#include<bits/stdc++.h>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);}

B

模拟一下，暴一暴就过了啊？

#include<bits/stdc++.h>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;}

C

继续模拟

#include<bits/stdc++.h>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<a.x;}};vector<edge> 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<DO[i].size();j++){at=DO[i][j].x;vector<edge> :: 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<edge> :: 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<UP[i].size();j++){at=UP[i][j].x;vector<edge> :: 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<edge> :: 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);}}

D

花式暴力？

#include<string>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>#include<set>#include<map>#include<queue>#include<vector>using namespace std;string a, b;struct cnt {int dt[30];void init(){memset(dt, 0, sizeof(dt));return;}inline bool operator<(const cnt & rig)const{for (int i = 0; i < 30; i++){if (dt[i] != rig.dt[i]){return dt[i] < rig.dt[i];}}return 0;}};int check(int x){cnt tmp;tmp.init();set<cnt>st;for (int i = 0; a[i]; i++){tmp.dt[a[i] - 'a']++;if (i >= x){tmp.dt[a[i - x] - 'a']--;}if (i >= x - 1){st.insert(tmp);}}tmp.init();for (int i = 0; b[i]; i++){tmp.dt[b[i] - 'a']++;if (i >= x){tmp.dt[b[i - x] - 'a']--;}if (i >= x - 1){if (st.find(tmp) != st.end()){return 1;}}}return 0;}int main(){std::ios::sync_with_stdio(false);cin >> a >> b;int tar = 0;for (int i = a.size(); i > 0; i--){if (check(i)){cout << i << "\n";tar = 1;break;}}if (!tar){cout << "0\n";}//cin >> a;return 0;}

E

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=0def 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 = cntdef 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)

F

bitset优化暴力

#include<bits/stdc++.h>#define xx first#define yy second#define mp make_pairusing namespace std;const int MAXN=1005;vector<int> sv[MAXN][MAXN];vector<pair<int,int> > edge[MAXN];bitset<MAXN> G[MAXN];map<string,int> 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<MAXN> 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<bits/stdc++.h>using namespace std;set<int> 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(x<now-1||sv.count(0))puts("No");else if(x==now-1&&sv.size()&&*(--sv.end())>x)puts("No");else{while(!sv.insert(x).second){sv.erase(x);x--;}while(sv.count(now))now++;puts("Yes");}}return 0;}

H

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<bits/stdc++.h>#define xx first#define yy second#define mp make_pairusing namespace std;const int MAXN=1e5+5;vector<pair<int,int> > sv[3];vector<int> e[MAXN],E[MAXN],et[MAXN];//Qint 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<int> 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<int> 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<Q.size();i++){int u=Q[i];for(auto v:E[u]){in[v]--;if(!in[v]){Q.push_back(v);}}}if(Q.size()!=tot){return 0*printf("Infinite\n");}for(auto i:sv[1]){int x,y;tie(x,y)=i;et[x].push_back(y);}for(auto i:Q){dfsD(i,0);dfsU(i,0);dfsT(i,0);}printf("%d\n",*max_element(dp+1,dp+1+n));return 0;}

I

这竟然，是个数学题啊？

#include<bits/stdc++.h>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(x<y)printf("0 1\n");elseprintf("1 0\n");}else{int z = gcd(x,y,x1,y1);int xx = x/z,yy = y/z;int ans = (xx+yy);if(ans<=50000){printf("4\n0 0\n%d %d\n%d %d\n%d %d\n",x-1,y,xx,yy,x,y-1);continue;}printf("3\n");printf("0 0\n");printf("%d %d\n",x,y); while(x1>0&&y1>0){x1 -= yy;y1 -= xx;}printf("%d %d\n",abs(y1),abs(x1));}}}

J

模拟退火暴一暴

#include <iostream>#include <cstdio>#include <cmath>#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 <ctime>#include<cstdlib>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<n; i++) ans+=cross_area(p[i],p[i+1],C);    return fabs(ans);}double x,y,v,det,t,g,R;const double pi = acos(-1.0); double X,Y;const double err = 1e-6;double Rand(double L, double R) {    return (rand() % 10000) / 10000.0 * (R - L) + L;  } int main() {    srand(time(0));scanf("%d%lf",&n,&R);C.R=R;for(int i=0; i<n; i++) scanf("%lf%lf",&p[i].x,&p[i].y);    C.p.x = 0;    C.p.y = 0;double anss = 0;    for(int i=0;i<n;i++)        anss = cal_cross(C,p,n);;    //printf("%lf\n",anss);double minx =1e5,miny = 1e5,maxx = 0,maxy = 0;minx = 0.0;miny = 0.0;maxx = 100.0;maxy = 100.0;    point cnt;    double acost = 0.0;for(int ii = 1;ii <= n;ii++){point ans;ans.x = p[ii].x;ans.y = p[ii].y;C.p = ans;   double ncost = cal_cross(C,p,n);    point tmp;    double step = max(maxx-minx,maxy-miny);    int flag = -1;    for (int i = 0;step > 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;}

K

奇奇怪怪的博弈。。。

#define _CRT_SECURE_NO_WARNINGS#include<iostream>#include<queue>#include<algorithm>#include<cstring>#include<cstdio>#include<cctype>#include<string>#include<cmath>#include<stack>#include<set>#include<map>#include<vector>#include<sstream>using namespace std;string data;long long cnt[50], tmps;int num[50];int n;long long base = (1LL << 40);long long ans;map<long long, long long>mp;int main(){cin >> n;for (int i = 0; i < n; i++){cin >> data;num[i] = data.size();int tar = 0;tmps = base;for (int j = 0; data[j]; j++){if (tar || (j != 0 && data[j] != data[j - 1])){tmps /= 2;tar = 1;}if (data[j] == 'B'){cnt[i] += tmps;}else{cnt[i] -= tmps;}}}int siz1 = n / 2;int siz2 = n - siz1;long long val, cos;for (long long i = 0; i < (1 << siz1); i++){cos = 0;val = 0;for (int j = 0; j < siz1; j++){if (i&(1LL << j)){cos += cnt[j];val += num[j];}}if (mp.find(cos) != mp.end()){mp[cos] = max(mp[cos], val);}else{mp[cos] = val;}}long long ans = 0;for (long long i = 0; i < (1 << siz2); i++){cos = 0;val = 0;for (int j = 0; j < siz2; j++){if (i&(1LL << j)){cos += cnt[j + siz1];val += num[j + siz1];}}if (mp.find(-1*cos) != mp.end()){ans = max(ans, mp[-1 * cos] + val);}}cout << ans << "\n";//cin >> n;return 0;}