2014 Asia Beijing Regional Contest

A. HDU 5112

//      whn6325689#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath>#include <functional>#include <numeric>using namespace std;typedef long long ll;typedef long double ld;typedef pair<ll, ll> pll;typedef complex<ld> point;typedef pair<int, int> pii;typedef pair<pii, int> piii;typedef vector<int> vi;#define CLR(x,y) memset(x,y,sizeof(x))#define mp(x,y) make_pair(x,y)#define pb(x) push_back(x)#define lowbit(x) (x&(-x))#define MID(x,y) (x+((y-x)>>1))#define eps 1e-9#define INF 0x3f3f3f3f#define LLINF 1LL<<62template<class T>inline bool read(T &n){    T x = 0, tmp = 1; char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;}template <class T>inline void write(T n){    if(n < 0)    {        putchar('-');        n = -n;    }    int len = 0,data[20];    while(n)    {        data[len++] = n%10;        n /= 10;    }    if(!len) data[len++] = 0;    while(len--) putchar(data[len]+48);}//-----------------------------------struct Node{double a,b;}t[10010];bool cmp(Node a,Node b){return a.a<b.a;}int main(){int T,n,m,cas=0;read(T);while(T--){read(n);for(int i=0;i<n;i++)scanf("%lf %lf",&t[i].a,&t[i].b);sort(t,t+n,cmp);double ans=-INF;for(int i=1;i<n;i++){double tt=t[i].a-t[i-1].a;double dis=abs(t[i].b-t[i-1].b);double v=dis/tt;if(v>ans)ans=v ;}printf("Case #%d: %.2f\n",++cas,ans); }return 0;}

B. HDU 5113

暴力染色

多用几种染色的方法，然后check一下

多种之后不行就是不行咯

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#define maxn 10#define maxk 100using namespace std;int t,n,m,k;int a[maxn][maxn];struct C{    int id;    int num;    bool operator <(const C &x) const    {        return num<x.num;    }}c[maxk];bool check(){    for(int i=1;i<=n;i++)        for(int j=1;j<=m;j++)        if(a[i][j] == a[i][j-1] || a[i][j] == a[i][j+1] || a[i][j] == a[i-1][j] || a[i][j] == a[i+1][j]) return false;    return true;}void print(){    printf("YES\n");    for(int i=1;i<=n;i++)        for(int j=1;j<=m;j++)        printf("%d%c",a[i][j],j==m?'\n':' ');}void solve1(){    memset(a,0,sizeof(a));    int Index = 1;    C temp;    temp = c[Index];int num=0;int b[maxn][maxn];    for(int i=1;i<=n;i++)        for(int j=i%2==0?2:1;j<=m;j+=2)    {        if(temp.num == 0) {Index++;temp = c[Index];}        b[i][j] = num++;        a[i][j] = c[Index].id;        temp.num--;    }    for(int i=1;i<=n;i++)        for(int j=(i+1)%2==0?2:1;j<=m;j+=2)    {        if(temp.num == 0) {Index++;temp = c[Index];}        b[i][j] = num++;        a[i][j] = c[Index].id;        temp.num--;    }    for(int i=1;i<=n;i++){for(int j=1;j<=m;j++)cout<<b[i][j]<<" ";cout<<endl;}}void solve2(){    memset(a,0,sizeof(a));    int Index = 1;    C temp;    temp = c[Index];    for(int i=2;i<=m+n;i+=2)        for(int j=1;j<i;j++)        if(i-j>=1 && i-j<=n && j<=m)        {            if(temp.num == 0) {Index++;temp = c[Index];}            a[i-j][j] = c[Index].id;            temp.num--;        }    for(int i=3;i<=m+n;i+=2)        for(int j=1;j<i;j++)        if(i-j>=1 && i-j<=n && j<=m)        {            if(temp.num == 0) {Index++;temp = c[Index];}            a[i-j][j] = c[Index].id;            temp.num--;        }}void solve3(){    memset(a,0,sizeof(a));    int Index = 1;    C temp;    temp = c[Index];    for(int i=3;i<=m+n;i+=2)        for(int j=1;j<i;j++)        if(i-j>=1 && i-j<=n && j>=1 && j<=m)        {            if(temp.num == 0) {Index++;temp = c[Index];}            a[i-j][j] = c[Index].id;            temp.num--;        }    for(int i=2;i<=m+n;i+=2)        for(int j=1;j<i;j++)        if(i-j>=1 && i-j<=n && j>=1 && j<=m)        {            if(temp.num == 0) {Index++;temp = c[Index];}            a[i-j][j] = c[Index].id;            temp.num--;        }}int main(){    scanf("%d",&t);    for(int cas = 1;cas <=t;cas++)    {        scanf("%d%d%d",&n,&m,&k);        for(int i=1;i<=k;i++)        {            scanf("%d",&c[i].num);            c[i].id = i;        }printf("Case #%d:\n",cas);        sort(c+1,c+1+k);        reverse(c+1,c+1+k);        solve1();        if(check()) {print();continue;}        solve2();        if(check()) {print();continue;}        solve3();        if(check()) {print();continue;}        printf("NO\n");    }}

D. HDU 5114

简单的区间DP，廖神

//By LH#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;typedef long long LL;int n, a[205], b[205];LL f[205][205];LL dp(int x, int y){    int i;    LL tmp, now;    if (f[x][y] != -1) return f[x][y];    if (x+1 == y) return 0;    tmp = 1000000000;    for (i = x+1; i < y; i++)    {        now = dp(x, i) + dp(i, y) + a[i]+b[x]+b[y];        tmp = min(now, tmp);    }    f[x][y] = tmp;    return f[x][y];}int main(){    int i, T, cnt;    scanf("%d", &T);    cnt = 0;    while (T--)    {        cnt++;        scanf("%d", &n);        for (i = 1; i <= n; i++) scanf("%d", &a[i]);        a[0] = 0; a[n+1] = 0;        for (i = 1; i <= n; i++) scanf("%d", &b[i]);        b[0] = 0; b[n+1] = 0;        memset(f, -1, sizeof(f));        printf("Case #%d: %I64d\n", cnt, dp(0, n+1));    }    return 0;}

H. HDU 5119

简单的背包DP

如果数量级再大一些用异或方程组，见前几篇博客

//By LH#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;typedef long long LL;LL dp[2][1 << 22];  int num[44];    int main()  {      int t, n, m, icase = 1;      LL ans;      scanf("%d", &t);      while (t--)      {          memset (dp, 0, sizeof(dp));          scanf("%d%d", &n, &m);          for (int i = 1; i <= n; i++)                scanf("%d", &num[i]);          dp[0][0] = 1;          for (int i = 1; i <= n; i++)              for (int j = 0; j <= (1 << 20); j++)                  dp[i % 2][j] = dp[1 - (i % 2)][j] + dp[1 - (i % 2)][j ^ num[i]];          ans = 0;          for (int i = m; i <= (1 << 20); i++)              ans += dp[n % 2][i];            printf("Case #%d: %I64d\n", icase++, ans);      }      return 0;  }  

I. HDU 5120

简单的计算几何

大圆面积并 - 两个大圆小圆面积并 + 两个小圆并

A神代码

#include <cstdio>#include <iostream>#include <cmath>#include <cstring>#include <vector>#include <algorithm>#define eps 1e-8#define maxn 1010#define PI (acos(-1))using namespace std;int dlcmp(double x){    return x<-eps?-1:x>eps;}double sqr(double x){    return x*x;}struct Point{    double x,y;    Point(){};    Point(double xx,double yy):x(xx),y(yy){};    Point operator - (const Point &a) const {return Point(x-a.x,y-a.y);}    Point operator + (const Point &a) const {return Point(x+a.x,y+a.y);}    //double operator * (const Point &a) const {return x*a.x+y*a.y;}    Point trunc(double d)    {        double dis(Point,Point);        double len = dis(*this,Point(0,0));        return Point(x*d/len,y*d/len);    }    Point rotate(double a)    {        return Point( x*cos(a)-y*sin(a),y*cos(a)+x*sin(a) );    }};double dot(Point a,Point b){    return a.x*b.x+a.y*b.y;}struct Segment{    Point s,e;    Segment(){}    Segment(Point ss,Point ee):s(ss),e(ee){}};struct Rectangle{    Point lt;    Point rb;};struct Circle{    Point o;    double r;    Circle(){}    Circle(Point a,double l):o(a),r(l){}    double area(){return sqr(r)*PI;}};double cross(double x1,double y1,double x2,double y2){    return x1*y2-x2*y1;}double cross(Point a,Point b){    return a.x*b.y-b.x*a.y;}double cross(Point a,Point b,Point s){    double x1=a.x-s.x,y1=a.y-s.y;    double x2=b.x-s.x,y2=b.y-s.y;    return x1*y2-x2*y1;}double dis(Point a,Point b){    return sqrt( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y) );}int inter_circle(Circle c1,Circle c2,Point &p1,Point &p2){    double len=dis(c1.o,c2.o);    if( dlcmp(len - c1.r - c2.r)>0 ) return 0;    double s = ( sqr(c1.r)-sqr(c2.r)+sqr(len) )/len/2;    double h = sqrt( sqr(c1.r) - sqr(s) );    Point vec=c2.o-c1.o;    Point p0 = c1.o+vec.trunc(s);    p1 = p0+vec.rotate(PI/2).trunc(h);    p2 = p0-vec.rotate(PI/2).trunc(h);    return 1;}double vec_angle(Point a,Point b){    double tmp=dot(a,b)/(dis( a,Point(0,0) )*dis( b,Point(0,0) ));    if(dlcmp(tmp-1)>=0) tmp = 1;    if(dlcmp(tmp+1)<=0) tmp = -1;    return acos(tmp);}double arc_area(Point a,Point b,Circle c){    double theta = vec_angle(a-c.o,b-c.o);    double sf = sqr(c.r)*theta/2.0;    double st = sqr(c.r)*sin(theta)/2.0;    if(dlcmp(cross(a,b,c.o))>0)        return sf-st;    else        return c.area()-sf+st;}double circle_and(Circle a,Circle b){    if( dlcmp(dis(a.o,b.o) - a.r - b.r) >=0 ) return 0;    if(a.r>b.r) swap(a,b);    if(dlcmp(b.r - dis(a.o,b.o) - a.r)>=0) return a.area();    Point p1,p2;    inter_circle(a,b,p1,p2);    return abs( arc_area(p2,p1,a) + arc_area(p1,p2,b) );}int t,cas;double R,r;int main(){    scanf("%d",&t);    for(int cas = 1;cas <= t;cas++)    {        Circle a1,a2,b1,b2;        scanf("%lf%lf",&r,&R);        scanf("%lf%lf",&a1.o.x,&a1.o.y);        scanf("%lf%lf",&b1.o.x,&b1.o.y);        a2.o = a1.o;        b2.o = b1.o;        a1.r = r;        a2.r = R;        b1.r = r;        b2.r = R;        //cout<<circle_and(a2,b2)<<endl;        double ans = circle_and(a2,b2) - circle_and(a2,b1) - ( circle_and(a1,b2) - circle_and(a1,b1) );        printf("Case #%d: %.6lf\n",cas,ans);    }    return 0;}

K. HDU 5126

很简单的题目，但是一定要思考清楚

//      whn6325689#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath>#include <functional>#include <numeric>using namespace std;typedef long long ll;typedef long double ld;typedef pair<ll, ll> pll;typedef complex<ld> point;typedef pair<int, int> pii;typedef pair<pii, int> piii;typedef vector<int> vi;#define CLR(x,y) memset(x,y,sizeof(x))#define mp(x,y) make_pair(x,y)#define pb(x) push_back(x)#define lowbit(x) (x&(-x))#define MID(x,y) (x+((y-x)>>1))#define eps 1e-9#define INF 0x3f3f3f3f#define LLINF 1LL<<62template<class T>inline bool read(T &n){    T x = 0, tmp = 1; char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;}template <class T>inline void write(T n){    if(n < 0)    {        putchar('-');        n = -n;    }    int len = 0,data[20];    while(n)    {        data[len++] = n%10;        n /= 10;    }    if(!len) data[len++] = 0;    while(len--) putchar(data[len]+48);}//-----------------------------------int a[1000010];  int main()  {      int T,n,i,j;      int min,cnt,cas=1;      read(T);      while(T--)      {          cnt=0;          read(n);          for(i=0;i<n;i++)              read(a[i]);  min=INF;for(int i=n-1;i>=0;i--)if(a[i]<min)min=a[i];elsecnt++;        printf("Case #%d: %d\n",cas++,cnt);      }      return 0;  }