标签：its hat 出现 min res throw problems his printf
Calculate the number of toys that land in each bin of a partitioned toy box.
Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They gave John a rectangular box to put his toys in, but John is rebellious and obeys his parents by simply throwing his toys into the box. All the toys get mixed up, and it is impossible for John to find his favorite toys.
John‘s parents came up with the following idea. They put cardboard partitions into the box. Even if John keeps throwing his toys into the box, at least toys that get thrown into different bins stay separated. The following diagram shows a top view of an example toy box.
For this problem, you are asked to determine how many toys fall into each partition as John throws them into the toy box.Input
Output
Sample Input
5 6 0 10 60 0 3 1 4 3 6 8 10 10 15 30 1 5 2 1 2 8 5 5 40 10 7 9 4 10 0 10 100 0 20 20 40 40 60 60 80 80 5 10 15 10 25 10 35 10 45 10 55 10 65 10 75 10 85 10 95 10 0
Sample Output
0: 2 1: 1 2: 1 3: 1 4: 0 5: 1 0: 2 1: 2 2: 2 3: 2 4: 2
Hint
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<stack> #include<map> #include<vector> #include<set> using namespace std; const int MAX=5e3+10; const double eps=1e-4; const double mod=1e9+7; #define INF 0x7fffffff #define ll long long #define edl putchar(‘\n‘) #define useit ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); #define FOR(i,a,b) for(int i=a;i<=b;i++) #define ROF(i,a,b) for(int i=a;i>=b;i--) #define mst(a) memset(a,0,sizeof(a)) #define mstn(a,n) memset(a,n,sizeof(a)) struct point{double x,y;}a[MAX][3],b; //bool cmp(const num &x, const num &y){return x.v>y.v;} double cross(point a,point b){return (a.x*b.y-a.y*b.x);} double dot(point a,point b){return (a.x*b.x+a.y*b.y);} double cro(point a,point b,point c){return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);} int main() { double x1,y1,x2,y2; int n,m,cas=0; int c[MAX]; while(scanf("%d",&n)&&n) { scanf("%d%lf%lf%lf%lf",&m,&x1,&y1,&x2,&y2); if(cas) edl; mst(c); FOR(i,1,n)cin>>a[i][1].x>>a[i][2].x,a[i][1].y=y1,a[i][2].y=y2; //a[0][1].x=x1,a[0][1].y=y1,a[0][2].x=x1,a[0][2].y=y2; a[n+1][1].x=x2,a[n+1][1].y=y1,a[n+1][2].x=x2,a[n+1][2].y=y2; while(m--) { cin>>b.x>>b.y; FOR(i,1,n+1) { if(cro(a[i][2],a[i][1],b)>0) { c[i-1]++; break; } } } FOR(i,0,n) cout<<i<<": "<<c[i]<<endl; cas++; } }
至于第二题，题面几乎完全相同，不同点有二：一、线的给出不按从左到右，但仍不相交。二、答案要求略有不同
解决线的排序问题很简单，由于不会相交，只要按线的中点坐标排序即可。
代码如下：
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<stack> #include<map> #include<vector> #include<set> using namespace std; const int MAX=5e3+10; const double eps=1e-4; const double mod=1e9+7; #define INF 0x7fffffff #define ll long long #define edl putchar(‘\n‘) #define useit ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); #define FOR(i,a,b) for(int i=a;i<=b;i++) #define ROF(i,a,b) for(int i=a;i>=b;i--) #define mst(a) memset(a,0,sizeof(a)) #define mstn(a,n) memset(a,n,sizeof(a)) struct point{double x1,y1,x2,y2,c;}a[MAX],b; bool cmp(const point &x, const point &y){return x.c<y.c;} //double cross(point a,point b){return (a.x*b.y-a.y*b.x);} //double dot(point a,point b){return (a.x*b.x+a.y*b.y);} double cro(point a,point c){return (a.x2-c.x1)*(a.y1-c.y1)-(a.x1-c.x1)*(a.y2-c.y1);} int main() { double x1,y1,x2,y2; int n,m,cas=0; int c[MAX],d[MAX],e[MAX]; while(scanf("%d",&n)&&n) { scanf("%d%lf%lf%lf%lf",&m,&x1,&y1,&x2,&y2); if(cas) edl; mst(c); mst(d); FOR(i,1,n)cin>>a[i].x1>>a[i].x2,a[i].y1=y1,a[i].y2=y2,a[i].c=a[i].x1+a[i].x2; sort(a+1,a+n+1,cmp); //a[0][1].x=x1,a[0][1].y=y1,a[0][2].x=x1,a[0][2].y=y2; a[n+1].x1=x2,a[n+1].y1=y1,a[n+1].x2=x2,a[n+1].y2=y2; while(m--) { cin>>b.x1>>b.y1; FOR(i,1,n+1) { if(cro(a[i],b)>0) { c[i-1]++; break; } } } sort(c,c+n+1); printf("Box\n"); int jishu=1; FOR(i,0,n) { if(c[i]!=0) { if(c[i]==c[i+1]) jishu++; else { printf("%d: %d\n",c[i],jishu); jishu=1; } } } cas++; } }
标签：its hat 出现 min res throw problems his printf
原文地址：http://www.cnblogs.com/qq936584671/p/7732005.html
