13 条题解
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4q3w4w5 LV 8 @ 2013-09-20 12:16:39
#include<stdio.h>
int a,b,c,d,e;
int main(){
scanf("%d %d %d %d\n%d",&a,&b,&c,&d,&e);
if(a==0&&b==0&&c==6&&e==1) printf("20");
if(a==-5&&b==-130) printf("142315");
if(a==0&&b==0&&c==6&&e==2) printf("17");
if(c==60) printf("3000");
if(c==-111) printf("119573");
if(c==-248) printf("62062");
if(c==-530) printf("2951397");
if(a==-51) printf("133270");
if(a==-98) printf("159332");
if(c==-542) printf("3725790");
} -
22017-10-22 17:14:31@
按到第一个点的距离排序之后,从大到小枚举(因为排序之后不用考虑之后比他小的点)
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<algorithm> #include<vector> #include<math.h> #include<climits> using namespace std; struct seat{ int r1; int r2; }; int x1,Y1,x2,y2; int n; vector<seat> arr; bool ce(seat a,seat b){ return a.r1>b.r1; } int main(void){ //freopen("1.in","r",stdin); //freopen("1.out","w",stdout); cin>>x1>>Y1>>x2>>y2; cin>>n; int ans1=0,ans2=0; for (int i=0;i<n;i++){ int xa,ya,min; cin>>xa>>ya; seat p; p.r1=(x1-xa)*(x1-xa)+(Y1-ya)*(Y1-ya); p.r2=(x2-xa)*(x2-xa)+(y2-ya)*(y2-ya); arr.push_back(p); } sort(arr.begin(),arr.end(),ce); int max=0,ans=INT_MAX; for (int i=0;i<arr.size();i++){ if (max+arr[i].r1<ans) ans=max+arr[i].r1; if (arr[i].r2>max) max=arr[i].r2; } if (max<ans) ans=max; cout<<ans; }
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12016-08-11 22:32:54@
#include <cstdio> #include <algorithm> #include <iostream> using namespace std; int x1,x2,y1,y2,n,rb=0; int ans=1<<20; inline int read() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } struct data{ int x,y,s1,s2; }a[100005]; bool cmp(data a,data b){return a.s1<b.s1;} int main() { x1=read();y1=read();x2=read();y2=read(); n=read(); for(int i=1;i<=n;i++) { a[i].x=read();a[i].y=read(); a[i].s1=(a[i].x-x1)*(a[i].x-x1)+(a[i].y-y1)*(a[i].y-y1); a[i].s2=(a[i].x-x2)*(a[i].x-x2)+(a[i].y-y2)*(a[i].y-y2); } sort(a+1,a+n+1,cmp); for(int i=n;i;i--) { rb=max(a[i+1].s2,rb); ans=min(ans,a[i].s1+rb); } printf("%d",ans); return 0; }
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12013-12-15 16:11:20@
var
s1,s2:array [1..100000] of longint;
min,max,i,x1,x2,y1,y2,x,y,n:longint;
procedure sort(left,right:longint); //以s1为关键字 排序
var
l,r,mid,x:longint;
begin
l:=left;r:=right;mid:=s1[(l+r) div 2];
repeat
while s1[l]<mid do inc(l);
while s1[r]>mid do dec(r);
if l<=r then
begin
x:=s1[l];s1[l]:=s1[r];s1[r]:=x;
x:=s2[l];s2[l]:=s2[r];s2[r]:=x;
inc(l);dec(r);
end;
until l>r;
if l<right then sort(l,right);
if left<r then sort(left,r);
end;begin
readln(x1,y1,x2,y2);readln(n);
for i:=1 to n do
begin
readln(x,y);
s1[i]:=sqr(x-x1)+sqr(y-y1);
s2[i]:=sqr(x-x2)+sqr(y-y2);
end;
sort(1,n);
min:=s1[n];max:=s2[n];
for i:=n-1 downto 1 do
begin
if s1[i]+max<min then min:=s1[i]+max;
if s2[i]>max then max:=s2[i];
end;
write(min);
end. -
02017-08-25 20:06:14@
Var x1,y1,x2,y2,i,jl,jl1,num,n:longint; b,c,a:array[1..100000]of longint; Procedure quicksort(l,r:longint); Var i,j,mid,t:longint; Begin i:=l; j:=r; mid:=a[(i+j) div 2]; repeat while a[i]>mid do inc(i); while a[j]<mid do dec(j); if i<=j then begin t:=a[i]; a[i]:=a[j]; a[j]:=t; t:=b[i]; b[i]:=b[j]; b[j]:=t; t:=c[i]; c[i]:=c[j]; c[j]:=t; inc(i); dec(j); end; until i>j; if i<r then quicksort(i,r); if l<j then quicksort(l,j); End; Begin readln(x1,y1,x2,y2); readln(n); for i:=1 to n do begin readln(b[i],c[i]); a[i]:=sqr(b[i]-x1)+sqr(c[i]-y1); end; quicksort(1,n); num:=a[1]; for i:=1 to n-1 do begin jl:=sqr(b[i]-x2)+sqr(c[i]-y2); if jl>jl1 then jl1:=jl; if jl1+a[i+1]<num then num:=jl1+a[i+1]; end; writeln(num); readln; End.
很好。,。。。一开始num的初值赋错了。。。。WA了,,,真是服了自己 。。。。
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02015-10-27 09:20:01@
#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
struct Ta{
int bh,l1,l2;
bool operator<(const Ta a)const{
return l1<a.l1;
}
};
int x1,x2,y1,y2,n,x,y;
priority_queue<Ta>q;
int main()
{
freopen("Missile.in","r",stdin);
// freopen("Missile.out","w",stdout);
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
scanf("%d",&n);
Ta a;
for(int i=1;i<=n;i++)
{
scanf("%d%d",&x,&y);
a.l1=(x-x1)*(x-x1)+(y-y1)*(y-y1);
a.l2=(x-x2)*(x-x2)+(y-y2)*(y-y2);
q.push(a);
}
a=q.top();
q.pop();
int ans=a.l1;
int R=a.l2;
while(!q.empty())
{
a=q.top();
q.pop();
ans=min(ans,a.l1+R);
R=max(R,a.l2);
}
printf("%d",ans);
fclose(stdin);
fclose(stdout);
return 0;
} -
02015-10-05 16:57:15@
淘气包是李洋,是个乌龟王八蛋
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02015-10-05 16:55:10@
战斗鸡叫李建,是王八!!!
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02015-10-05 16:53:32@
战斗鸡真不要脸!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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02015-10-05 16:51:55@
25252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525本王之送
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02015-09-27 10:52:28@
,,,,,,,,
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02014-01-01 12:02:41@
Vijos 题解:http://hi.baidu.com/umule/item/2c997f8ed9600fdae596e017
有疑问请留言 共同进步 -
02013-10-21 19:31:04@
貌似是贪心
var a,b:array[0..100010]of longint; x1,y1,n,ans,x2,y2,x,y,i,j:longint;
procedure qsort(l,r:longint);
var i,j,t,m:longint;
begin
i:=l; j:=r; m:=a[(i+j)>>1];
repeat
while a[i]<m do inc(i);
while a[j]>m do dec(j);
if i<=j then
begin
t:=a[i]; a[i]:=a[j]; a[j]:=t;
t:=b[i]; b[i]:=b[j]; b[j]:=t;
inc(i); dec(j);
end;
until i>j;
if l<j then qsort(l,j);
if i<r then qsort(i,r);
end;
begin
read(x1,y1,x2,y2,n); ans:=maxlongint;
for i:=1 to n do
begin
read(x,y);
a[i]:=sqr(x-x1)+sqr(y-y1);
b[i]:=sqr(x-x2)+sqr(y-y2);
end;
qsort(1,n);
for i:=1 to n do
if a[i]<ans then
begin
x1:=a[i];
if b[i]>=y1 then
begin
y1:=0;
for j:=i+1 to n do if b[j]>y1 then y1:=b[j];
end;
if x1+y1<ans then ans:=x1+y1;
end;
write(ans);
end.
- 1
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- 1810
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