ASC 04 题解

A

求最小割方案是否唯一

#include <iostream>#include <cmath>#include <algorithm>#include <cstdio>#include <cstring>#include <string>#include <vector>#include <map>#include <functional>#include <cstdlib>#include <queue>#include <stack>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())typedef long long ll;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} #define MAXN (4000*2+100)#define MAXM (60000*2+10)int n,k;class tar{public:    vi G[MAXN],G2[MAXN];    int pre[MAXN],lowlink[MAXN],sccno[MAXN],dfs_clock,scc_cnt;    stack<int> S;     void dfs(int u) {        pre[u] = lowlink[u] = ++dfs_clock;        S.push(u);        int sz=SI(G[u]);        Rep(i,sz) {            int v=G[u][i];            if (!pre[v]) {                dfs(v);                lowlink[u]=min(lowlink[u],lowlink[v]);            } else if (!sccno[v]) {                lowlink[u]=min(lowlink[u],pre[v]);            }         }         if (lowlink[u]==pre[u]) {            scc_cnt++;              while(1) {                int x=S.top();S.pop();                sccno[x]=scc_cnt;                if (x==u) break;              }         }           }     void find_scc(int n) {        dfs_clock = scc_cnt = 0;        MEM(sccno)         MEM(pre)        Rep(i,n) if (!pre[i]) dfs(i);    }    void mem(int n) {        Rep(i,n) G[i].clear(),G2[i].clear();    }}S2;class Max_flow  //dinic+当前弧优化   {    public:        int n,t;        int q[MAXN];        ll weight[MAXM];    int edge[MAXM],Next[MAXM],Pre[MAXN],size;        void addedge(int u,int v,int w)          {              edge[++size]=v;              weight[size]=w;              Next[size]=Pre[u];              Pre[u]=size;          }          void addedge2(int u,int v,int w){addedge(u,v,w),addedge(v,u,w);}         bool b[MAXN];        ll d[MAXN];        bool SPFA(int s,int t)          {              MEMI(d)        MEM(b)            d[q[1]=s]=0;b[s]=1;              int head=1,tail=1;              while (head<=tail)              {                  int now=q[head++];                  Forp(now)                  {                      int &v=edge[p];                      if (weight[p]&&!b[v])                      {                          d[v]=d[now]+1;                          b[v]=1,q[++tail]=v;                      }                  }                  }              return b[t];          }         int iter[MAXN];      int dfs(int x,ll f)      {          if (x==t) return f;          Forpiter(x)          {              int v=edge[p];              if (weight[p]&&d[x]<d[v])              {                    int nowflow=dfs(v,min(weight[p],f));                    if (nowflow)                    {                      weight[p]-=nowflow;                      weight[p^1]+=nowflow;                      return nowflow;                    }              }          }          return 0;      }      ll max_flow(int s,int t)      {          (*this).t=t;        ll flow=0;          while(SPFA(s,t))          {              For(i,n) iter[i]=Pre[i];              ll f;              while (f=dfs(s,INF))                  flow+=f;           }          return flow;      }       void mem(int n)        {            (*this).n=n;          size=1;            MEM(Pre)       }    void init(int n) {        For(i,n) {            Forp(i) {                int v=edge[p];                  // cout<<i<<' '<<v<<' '<<weight[p]<<endl;                if (!weight[p]) {                    // if (i<=n1&&v>n1&&v<=n1+m1) {                    //     ::b[i][v-n1]=1;                    //     // cout<<i<<' '<<v-n1<<endl;                    // }                }                else S2.G[i-1].pb(v-1),S2.G2[i-1].pb(p);            }        }    }   }S;  struct {    int u,v;}e[MAXM];int main() {    freopen("attack.in","r",stdin);    freopen("attack.out","w",stdout);    int n,m,s,t;    n=read(),m=read(),s=read(),t=read();    S.mem(n);    For(i,m) {        int u=read(),v=read(),w=read();        S.addedge2(u,v,w);        e[i].u=u,e[i].v=v;    }    ll p=S.max_flow(s,t);    S.init(n);    S2.find_scc(n);    s--,t--;    bool fl=0;    For(i,m) {        e[i].u--,e[i].v--;        if (S.weight[i*2]) ;        else if (S2.sccno[e[i].u]==S2.sccno[s] && S2.sccno[e[i].v]==S2.sccno[t]) ;        else if (S2.sccno[e[i].u]!=S2.sccno[e[i].v]) fl=1;    }    S2.mem(n);    puts(fl?"AMBIGUOUS":"UNIQUE");    return 0;}

C

已知n个圆（n<=50)，求平面被分割成了几份。
欧拉定理：点数-边数+面数=2（平面图连通）
注意这题对精度要求较高，求圆交点避开三角函数，用sqrt代替

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (0x3f3f3f3f)#define F (1000000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()#define gmax(a,b) a=max(a,b);#define gmin(a,b) a=min(a,b);//#define double __float128//#define ld __float128typedef long long ll;typedef long double ld;ld sqr(ld a){return a*a;}const double eps=1e-8;int dcmp(double x) {    if (max(x,-x)<eps) return 0; else return x<0 ? -1 : 1; }ld PI = 3.141592653589793238462643383;class P{public:    double x,y;    P(double x=0,double y=0):x(x),y(y){}    friend ld dis2(P A,P B){return sqr(A.x-B.x)+sqr(A.y-B.y);   }    friend ld Dot(P A,P B) {return A.x*B.x+A.y*B.y; }//    friend ld Length(P A) {return sqrt(Dot(A,A)); }//    friend ld Angle(P A,P B) {//        if (dcmp(Dot(A,A))==0||dcmp(Dot(B,B))==0||dcmp(Dot(A-B,A-B))==0) return 0;//        return acos(max((ld)-1.0, min((ld)1.0, Dot(A,B) / Length(A) / Length(B) )) ); //    }    P rotate90()const{ return{ -y, x }; }    friend P operator- (P A,P B) { return P(A.x-B.x,A.y-B.y); }    friend P operator+ (P A,P B) { return P(A.x+B.x,A.y+B.y); }    friend P operator* (P A,double p) { return P(A.x*p,A.y*p); }    friend P operator/ (P A,double p) { return P(A.x/p,A.y/p); }    friend bool operator< (const P& a,const P& b) {return dcmp(a.x-b.x)<0 ||(dcmp(a.x-b.x)==0&& dcmp(a.y-b.y)<0 );}}; P read_point() {    P a;    scanf("%lf%lf",&a.x,&a.y);    return a;   } bool operator==(const P& a,const P& b) {    return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y) == 0;} typedef P V;/*欧拉公式： V+F-E=2 V-点数 F面数 E边数 */double Cos(double x) {    return 1-x*x/2+x*x*x*x/4/3/2-x*x*x*x*x*x/6/5/4/3/2;}double Sin(double x) {    return x/1-x*x*x/3/2+x*x*x*x*x/5/4/3/2/1;}struct C{    P c;    double r,x,y;    C(){}    C(P c,double r=0):c(c),r(r),x(c.x),y(c.y){}    P point(double a) {        return P(c.x+Cos(a)*r,c.y+Sin(a)*r);    }};double angle(V v) {return atan2(v.y,v.x);}int getCircleCircleIntersection(C C1,C C2,vector<P>& sol) {    double d = sqrt(dis2(C1.c,C2.c));    if (dcmp(d)==0) {        if (dcmp(C1.r - C2.r)==0) return -1; //2圆重合         return 0;    }    if (dcmp(C1.r+C2.r-d)<0) return 0;    if (dcmp(max(C2.r-C1.r,C1.r-C2.r)-d)>0) return 0;    d=dis2(C1.c,C2.c);    double t1 = (C1.r*C1.r - C2.r*C2.r) / (2 * d) + 0.5;    double t2 = C1.r*C1.r / d - t1*t1;    t2 = (dcmp(t2)<=0) ? (0): sqrt(max(0.0,t2));    P foot = C1.c + (C2.c - C1.c)*t1;    P p1 = foot + (C2.c - C1.c).rotate90() * t2;    P p2 = foot * 2 - p1;    sol.pb(p1);    if (p1==p2) return 1;    sol.pb(p2);    return 2; }//void circleCircleIntersectPoint(const Point& o1, double r1, const Point& o2, double r2, Point& ret1, Point& ret2)//{//  double d2 = o1.dis2(o2);//  double t1 = (r1*r1 - r2*r2) / (2 * d2) + 0.5;//  double t2 = r1*r1 / d2 - t1*t1;//  t2 = LEQ(t2, 0) ? 0 : sqrt(t2);//  Point foot = o1 + (o2 - o1)*t1;//  ret1 = foot + (o2 - o1).rotate90()*t2;//  ret2 = foot * 2 - ret1;//}int n;#define MAXN (102345)class bingchaji{public:    int father[MAXN],n,cnt;    void mem(int _n)    {        n=cnt=_n;        For(i,n) father[i]=i;    }    int getfather(int x)     {        if (father[x]==x) return x;        return father[x]=getfather(father[x]);    }    void unite(int x,int y)    {        x=getfather(x);        y=getfather(y);        if (x^y) {            --cnt;            father[x]=y;        }    }    bool same(int x,int y)    {        return getfather(x)==getfather(y);    }}S;void print(double a) {    printf("%.6lf",a); }void print(P p) {    printf("(%.6lf,%.6lf)",p.x,p.y);}template<class T>void print(vector<T> v) {    sort(v.begin(),v.end());    putchar('[');    int n=v.size();    Rep(i,n) {        print(v[i]);        if (i<n-1) putchar(',');     }    puts("]");}void CircleUnion(C c[],int n) {    ll dotsum=0,edgesum=0;     vector<P> Sol;    S.mem(n);    Rep(i,n) {        int cnt = 0;        vector<P> sol; sol.clear();        Rep(j,n) if (i^j) {            int sz=SI(sol);            getCircleCircleIntersection(c[i],c[j],sol);            getCircleCircleIntersection(c[i],c[j],Sol);            sz=SI(sol)-sz;            if (sz) {                S.unite(i+1,j+1);            }        }        sort(ALL(sol));        sol.erase(unique(ALL(sol)),sol.end());        edgesum+=SI(sol);    }    sort(ALL(Sol));    Sol.erase(unique(ALL(Sol)),Sol.end());    dotsum=SI(Sol);    int p=S.cnt;    cout<<-dotsum+edgesum+2*p-p+1<<endl;}C cir[MAXN];int main(){  freopen("circles.in","r",stdin);  freopen("circles.out","w",stdout);//  freopen("C.in","r",stdin);    while(cin>>n) {        Rep(i,n) {            ll x,y,r;            cin>>x>>y>>r;            cir[i].x=x,cir[i].y=y,cir[i].r=r;            cir[i].c=P(x,y);        }        CircleUnion(cir,n);    }    return 0;}

D

输入一个单纯形问题，按指定格式输出它的对偶问题。

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (0x3f3f3f3f)#define F (1000000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()#define gmax(a,b) a=max(a,b);#define gmin(a,b) a=min(a,b);typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}inline int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} #define MAXN (210)char s[10000];bool is_min;int a[210],n,m;int sig[MAXN];isdig(char c) {    return isdigit(c)|c=='+'|c=='-';} int getint(int &i) {    int x=0,f=1;    if (s[i]=='x') {        ++i;return 1;    }    if (!isdigit(s[i])) {        if (s[i]=='+') i++;        else if (s[i]=='-') f=-1,++i;    }    bool fl=0;    while(isdigit(s[i])) {        x=x*10+s[i]-'0'; ++i; fl=1;    }    while(!isdig(s[i]) &&s[i]) ++i;    if (!x&&!fl) x=1;    return x*f;}int f[MAXN][MAXN]={};void get(vi &v) {    int l=strlen(s);    for(int i=0;i<l;) {        int p=getint(i);        v.pb(p);    }}void get1() {    vi v; MEM(a)    get(v);    for(int i=0;i<SI(v);i+=2) {        a[v[i+1]]=v[i];    }}int sig2[MAXN];void getsig(int &si) {    char *c=strchr(s,'>');    if (NULL!=strchr(s,'>')) si=1;    else if (NULL!=strchr(s,'<')) si=-1;    else si=0;}int p[MAXN];void pri(int a[]=p) {    bool fl=0;    For(i,m) if (a[i]) {        if (a[i]==1) {            if (fl) putchar('+');        }        else if (a[i]==-1) putchar('-');        else {            if (fl&&a[i]>0) putchar('+');            cout<<a[i];        }        putchar('y');        cout<<i;        fl=1;    }    if(!fl) putchar('0');}int c[MAXN];int main(){//  freopen("D.in","r",stdin);    freopen("dual.in","r",stdin);    freopen("dual.out","w",stdout);    scanf("%d %d\n",&n,&m);    scanf("%s",s);    if (strcmp(s,"min")==0) is_min=1;else is_min=0;    scanf("%s\n",s);    get1();    scanf("%s\n",s);    For(i,n) {        cin.getline(s,5000);        getsig(sig[i]);    }    scanf("%s\n",s);    For(i,m) {        cin.getline(s,5000);        getsig(sig2[i]);        vi v;        get(v);        int sz=SI(v);        for(int j=0;j<sz-2;j+=2) {            f[i][v[j+1]]=v[j];        }        c[i]=v[sz-1];//      Rep(i,sz) cout<<v[i] <<' ';puts("");    }//  PRi2D(f,m,n)    printf("%d %d\n",m,n);    if (is_min) cout<<"max ";else cout<<"min ";//  PRi(c,m)/    pri(c);    puts("");    puts("with");    For(i,m) {        if (is_min) {            if (sig2[i]==1) printf("y%d>=0\n",i);            else if (sig2[i]==-1) printf("y%d<=0\n",i);            else printf("y%d arbitary\n",i);        } else {            if (sig2[i]==-1) printf("y%d>=0\n",i);            else if (sig2[i]==1) printf("y%d<=0\n",i);            else printf("y%d arbitary\n",i);        }    }    puts("under");    For(i,n) {        For(j,m) p[j]=f[j][i];        pri();        if (!is_min) {            if (sig[i]==1) printf(">=");            else if (sig[i]==-1) printf("<=");            else printf("=");         } else {            if (sig[i]==-1) printf(">=");            else if (sig[i]==1) printf("<=");            else printf("=");         }        cout<<a[i];        puts("");     }    return 0;}