ZOJ 2012 2月 月赛部分题解

临比赛前最后一篇博文.....T T......鸭梨山大的啊

C 训练赛的时候搞麻烦了，其实根本用不着数据结构。。。。省略代码

D 三点不共线降低了难度，根据基础的数学知识，如果画一条新的线段，与之前m条线段在区间内相交，则新生成m+1个part

O(n*n)超时，第一个反应就是看看排序，按照线段的起点坐标排序，发现之后是逆序数。。。附代码

#include 
#include 
#include 
#include 
#include 
#include 

#define cl(a) memset(a,0,sizeof(a))
#define ss(a) scanf("%d",&a)
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define N 30005
#define ll long long

using namespace std;

struct lining
{
    int k;
    int b;
    int st;
    int id;
    int num;
};

struct finalpoint
{
    int value;
    int id;
}c[N];

lining a[N];
int h[N],tree[N<<2];

bool cmp(lining p,lining q)
{
    return p.st> 1;
    if (i<=mid) update(i,lson);
    else update(i,rson);
    pushup(rt);
}

int query(int L,int R,int l,int r,int rt)
{
    if (L<=l&&r<=R)
    {
        return tree[rt];
    }
    int mid = (l+r) >> 1,res = 0;
    if (L<=mid) res+=query(L,R,lson);
    if (R>mid) res+=query(L,R,rson);
    return res;
}

int main()
{
    int i,j,x1,x2,n;
    while (ss(x1)!=EOF)
    {
        ss(x2);
        ss(n);
        for (i=1;i<=n;i++)
        {
            ss(a[i].k);ss(a[i].b);
            a[i].st = a[i].k*x1+a[i].b;
            a[i].id = i;
            c[i].value = a[i].k*x2 + a[i].b;
            c[i].id = i;
        }
        sort(a+1,a+n+1,cmp);
        sort(c+1,c+n+1,cmp1);
        cl(h);
        for (i=1;i<=n;i++) h[c[i].id] = i;
        for (i=1;i<=n;i++)
            a[i].num = h[a[i].id];
        cl(tree);
        ll sum = 1;
        for (i=1;i<=n;i++)
        {
            ll t= a[i].num , qy = 0;
            if (t > 1 ) qy = query(1,t-1,1,n,1);
            sum += (t-qy);
            update(t,1,n,1);
        }
        cout<

E 几何题，刚开始不会

后来才明白，在极坐标的环境下，压缩坐标倍数即可。椭圆变成了圆，半径为1.枚举所有可能为结果的圆，即过任意两个点的圆（不在圆内），看看哪个圆覆盖面积最大

#include 
#include 
#include 

#define N 220

using namespace std;

struct point
{
    double x;
    double y;
}u[N];

double dist(point A,point B)
{
    return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));
}

bool creatcircle(point &O, point A,point B)
{
    double l = dist(A,B);
    if (l>2.0) return 0;
    point C;
    C.x = (A.x + B.x) / 2.0;
    C.y = (A.y + B.y) / 2.0;
    double l1 = sqrt (1 - (l/2.0)*(l/2.0));
    double k = -1 * (A.x - B.x) / (A.y - B.y);
    O.x = C.x + l1 / sqrt(1+k*k);
    O.y = C.y + l1 * k / sqrt(1+k*k);
    return 1;
}

int check(point O,int n)
{
    int i,sum = 0;
    for (i=1;i<=n;i++)
    {
        if (dist(u[i],O)<=1) sum++;
    }
    return sum;
}

int main()
{
    int i,j,n;
    double a,b;
    while (cin>>a>>b)
    {
        cin>>n;
        for (i=1;i<=n;i++)
        {
            cin>>u[i].x>>u[i].y;
            u[i].x/=a;u[i].y/=b;
        }
        int res = 1;
        for (i=2;i<=n;i++)
            for (j=1;j

H 可以用DP做，DP弱，没有想到。。。。仔细想想，主要原因是两个长方形相交时，一整个矩形全部更新，而不是交叉区域更新。代码显而易见

判断矩形的时候WA了好几次。。。这个题虽然难度不大，但挺不错的~

#include 
#include 
#include 
#include 
#include 
#include 

#define ss(a) scanf("%d",&a)
#define cl(a) memset(a,0,sizeof(a))
#define N 1005

using namespace std;

struct ractangle
{
    int x1;
    int y1;
    int x2;
    int y2;
    int h;
}a[N];

int f[N];

bool crossing(ractangle p,ractangle q)
{
    if (p.x1>=q.x2||q.x1>=p.x2) return 0;
    if (p.y1>=q.y2||q.y1>=p.y2) return 0;
    return 1;
}

bool cmp(ractangle p,ractangle q)
{
    return p.x1n?n:a[i].x1+dx;
            a[i].y2=(a[i].y1+dy)>m?m:a[i].y1+dy;
            a[i].h=height;
        }
        int res=0;
        for (i=1;i<=c;i++)
        {
            f[i] = 0;
            for (j=1;jf[i]?f[j]:f[i];
            }
            f[i] += a[i].h;
            res = f[i]>res?f[i]:res;
        }
        cout<

F , J等待更新.....