T1：Blue
Greedy on the end, apparently a clam try to jump away is optimal, with each stone skipping across a number of updating the answer would be finished.
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<cstdlib> 7 using namespace std; 8 const int MAXN=1000005,INF=0x3f3f3f3f; 9 int T; 10 int n,m,d,L,a[MAXN],ans; 11 inline int R() { 12 int a=0;char c=getchar(); 13 while(c>'9'||c<'0')c=getchar(); 14 while(c>='0'&&c<='9')a=a*10+c-'0',c=getchar(); 15 return a; 16 } 17 bool flag; 18 int main() { 19 T=R(); 20 while(T--) { 21 n=R();m=R();d=R();L=R(); 22 for(int i=1;i<=n;i++) a[i]=R(); 23 a[0]=0;a[n+1]=L;ans=m; 24 int now=0; 25 for(int i=0;i<=n+1;i++) { 26 while(now!=n+1&&a[now+1]-a[i]<=d) ++now; 27 if(i==now) { 28 printf("0\n"),flag=1; 29 break; 30 } 31 if(now==n+1) break; 32 ans=min(ans,now-i); 33 } 34 if(!flag) { 35 if(ans==m) printf("Excited\n"); 36 else printf("%d\n",ans); 37 } 38 flag=0; 39 } 40 return 0; 41 }
T2：Weed
Maintenance segment tree with a sequence of operations, difficulties how to write in pushup, considering each node maintains three messages:
Can delete layers, layers left, the rest of the height and
The first two information (1) is determined by a simple calculation $ $ O
The third information can be calculated recursively down, embodied ask to see the function code
#include<cstdio> #include<iostream> #include<cmath> #include<cstring> #include<algorithm> #include<queue> #include<vector> #include<cstdlib> using namespace std; const int MAXN=300005; int m,q; struct opt { int k,v; }o[MAXN]; struct node { int l,r,sum,c,e; }tr[MAXN*4]; int ask(int p,int e) { if(e==tr[p<<1|1].c) return tr[p].sum-tr[p<<1|1].sum; if(e<tr[p<<1|1].c) return tr[p].sum-tr[p<<1|1].sum+ask(p<<1|1,e); return ask(p<<1,e-tr[p<<1|1].c+tr[p<<1|1].e); } void up(int p) { if(tr[p<<1|1].e>=tr[p<<1].c) { tr[p].e=tr[p<<1].e+tr[p<<1|1].e-tr[p<<1].c; tr[p].c=tr[p<<1|1].c; tr[p].sum=tr[p<<1|1].sum; } else if(!tr[p<<1|1].e) { tr[p].c=tr[p<<1].c+tr[p<<1|1].c; tr[p].sum=tr[p<<1].sum+tr[p<<1|1].sum; tr[p].e=tr[p<<1].e; } else { tr[p].e=tr[p<<1].e; tr[p].c=tr[p<<1|1].c+tr[p<<1].c-tr[p<<1|1].e; tr[p].sum=tr[p<<1|1].sum+ask(p<<1,tr[p<<1|1].e); } } void build(int p,int l,int r) { tr[p].l=l,tr[p].r=r; if(l==r) { tr[p].sum=o[l].k?0:o[l].v; tr[p].c=o[l].k?0:1; tr[p].e=o[l].k?o[l].v:0; return; } int mid=(l+r)>>1; build(p<<1,l,mid); build(p<<1|1,mid+1,r); up(p); } void change(int p,int x,int k,int c) { if(tr[p].l==tr[p].r) { tr[p].sum=k?0:c; tr[p].c=k?0:1; tr[p].e=k?c:0; return; } int mid=(tr[p].l+tr[p].r)>>1; if(x<=mid) change(p<<1,x,k,c); else change(p<<1|1,x,k,c); up(p); } int main() { scanf("%d%d",&m,&q); for(int i=1;i<=m;i++) scanf("%d%d",&o[i].k,&o[i].v); build(1,1,m); for(int i=1,aa,bb,cc;i<=q;i++) { scanf("%d%d%d",&aa,&bb,&cc); change(1,aa,bb,cc); printf("%d\n",tr[1].sum); } return 0; }
ps: a similar problem - building reconstruction
T3：Drink
God title list
For each point (one can be seen), a predetermined forward direction (which side faces)
And recorded in this direction all around the points are those points
(To the person's perspective who he's around, respectively)
For once modified, the information need only modify the Master of square boundary
Note that no records not everyone's direction, because this would modify reduced complexity $ O (n ^ 2) $
Only record of an individual (such as the upper left corner) direction and number
Before and after the introduction of information about reuse other people's direction
1 #include<cmath> 2 #include<cstring> 3 #include<algorithm> 4 #include<cstring> 5 #include<iostream> 6 #include<cstdlib> 7 #include<queue> 8 #include<vector> 9 using namespace std; 10 const int MAXN=2005; 11 int n,m,q,s[MAXN][MAXN]; 12 /* 13 0 14 3 x 1 15 2 16 */ 17 struct Point { 18 int id; 19 Point() {id=0;} 20 int x() { 21 return (id-1)/m+1; 22 } 23 int y() { 24 return (id-1)%m+1; 25 } 26 }fp,tmp[8][MAXN]; 27 int fd,tmpd[8][MAXN]; 28 bool operator == (const Point &aa,const Point &bb) { 29 return aa.id==bb.id; 30 } 31 Point New(int _x,int _y) { 32 Point ret; 33 ret.id=(_x-1)*m+_y; 34 return ret; 35 } 36 struct Node { 37 Point p[4]; 38 }a[MAXN][MAXN]; 39 inline int my_d(int d,int opt) { 40 return (opt-d+4)%. 4 ; 41 is } 42 is Point get_p (Point p, int d, int opt) { 43 is // opt facing direction d of point p of the point 44 is return . A [PX ()] [Py ()] p [my_d (d , opt)]; 45 } 46 is int get_d (point lp, point p, int d, int opt) { 47 // direction opt facing direction d of point lp of the point p 48 int Topt = (opt + 2 )% . 4 ; 49 for ( int I = 0 ; I < . 4 ; I ++ ) 50 IF(get_p (P, I, Topt) == LP) return I; 51 is return - . 1 ; 52 is } 53 is void Move (Point & P, int & D, int opt) { 54 is Point TP = P; 55 P = get_p (P, d, opt); 56 is d = get_d (TP, p, d, opt); 57 is } 58 void Modify (point p, int d, point c) { 59 // points on the direction of the p d c to point 60 . A [PX ()] [Py ()] P [D] = C; 61 is } 62 is void Change (int x,int y,int c) { 63 Point np=fp; 64 int nd=fd; 65 for(int i=1;i<x;i++) move(np,nd,2); 66 for(int i=1;i<y;i++) move(np,nd,1); 67 for(int i=1;i<=c;move(np,nd,1),++i) { 68 tmp[0][i]=np,tmp[4][i]=get_p(np,nd,0); 69 tmpd[0][i]=nd,tmpd[4][i]=get_d(np,tmp[4][i],nd,0); 70 } 71 move(np,nd,3); 72 for(int i=1;i<=c;move(np,nd,2),++i) { 73 tmp[1][i]=np,tmp[5][i]=get_p(np,nd,1); 74 tmpd[1][i]=nd,tmpd[5][i]=get_d(np,tmp[5][i],nd,1); 75 } 76 move(np,nd,0); 77 for(int i=1;i<=c;move(np,nd,3),++i) { 78 tmp[2][i]=np,tmp[6][i]=get_p(np,nd,2); 79 tmpd[2][i]=nd,tmpd[6][i]=get_d(np,tmp[6][i],nd,2); 80 } 81 move(np,nd,1); 82 if(x==1&&y==1) fp=np,fd=(nd+1)%4; 83 for(int i=1;i<=c;move(np,nd,0),++i) { 84 tmp[3][i]=np,tmp[7][i]=get_p(np,nd,3); 85 tmpd[3][i]=nd,tmpd[7][i]=get_d(np,tmp[7][i],nd,3); 86 } 87 for(int i=0;i<4;i++) 88 for(int j=1;j<=c;j++) 89 modify(tmp[i][j],my_d(tmpd[i][j],i),tmp[(i+1)%4+4][j]); 90 for(int i=4;i<8;i++) 91 for(int j=1;j<=c;j++) 92 modify(tmp[i][j],my_d(tmpd[i][j],(i+2)%4),tmp[(i-1)%4][j]); 93 } 94 void print() { 95 Point xp=fp,yp; 96 int xd=fd,yd; 97 for(int i=1;i<=n;move(xp,xd,2),++i) { 98 yp=xp;yd=xd; 99 for(int j=1;j<=m;move(yp,yd,1),++j) 100 printf("%d ",s[yp.x()][yp.y()]); 101 printf("\n"); 102 } 103 } 104 int main() { 105 scanf("%d%d%d",&n,&m,&q); 106 for(int i=0;i<=n+1;i++) 107 for(int j=0;j<=m+1;j++) { 108 if(i!=0&&i!=n+1&&j!=0&&j!=m+1) 109 scanf("%d",&s[i][j]); 110 if(i!=0) a[i][j].p[0]=New(i-1,j); 111 if(i!=n+1) a[i][j].p[2]=New(i+1,j); 112 if(j!=0) a[i][j].p[3]=New(i,j-1); 113 if(j!=m+1) a[i][j].p[1]=New(i,j+1); 114 } 115 fp=New(1,1);fd=0; 116 for(int i=1,aa,bb,cc;i<=q;i++) { 117 scanf("%d%d%d",&aa,&bb,&cc); 118 change(aa,bb,cc); 119 } 120 print(); 121 return 0; 122 }