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HDU-4419-Colourful Rectangle(线段树)
Problem Description
We use Red, Green and Blue to make new colours. See the picture below:
Now give you n rectangles, the colour of them is red or green or blue. You have calculate the area of 7 different colour. (Note: A region may be covered by same colour several times, but it’s final colour depends on the kinds of different colour)
Now give you n rectangles, the colour of them is red or green or blue. You have calculate the area of 7 different colour. (Note: A region may be covered by same colour several times, but it’s final colour depends on the kinds of different colour)
Input
The first line is an integer T(T <= 10), the number of test cases. The first line of each case contains a integer n (0 < n <= 10000), the number of rectangles. Then n lines follows. Each line start with a letter C(R means Red, G means Green, B means Blue) and four integers x1, y1, x2, y2(0 <= x1 < x2 < 10^9, 0 <= y1 < y2 < 10^9), the left-bottom‘s coordinate and the right-top‘s coordinate of a rectangle.
Output
For each case, output a line "Case a:", a is the case number starting from 1,then 7 lines, each line contain a integer, the area of each colour. (Note: You should print the areas as the order: R, G, B, RG, RB, GB, RGB).
Sample Input
3 2 R 0 0 2 2 G 1 1 3 3 3 R 0 0 4 4 G 2 0 6 4 B 0 2 6 6 3 G 2 0 3 8 G 1 0 6 1 B 4 2 7 7
Sample Output
Case 1: 3 3 0 1 0 0 0 Case 2: 4 4 12 4 4 4 4 Case 3: 0 12 15 0 0 0 0
Source
2012 ACM/ICPC Asia Regional Hangzhou Online
思路:维护区间R,G,B三种颜色的数量,更新时判断区间的组合颜色,再考虑与左右儿子的颜色组合的情况。
#include <cstdio> #include <map> #include <cmath> #include <algorithm> using namespace std; map<int,int>ii; struct S{ int r,g,b; int cr,cg,cb,crg,crb,cgb,crgb; }node[200000]; struct L{ int pos,l,r,type,flag; bool operator<(const L &p) const { return pos<p.pos; } }line[20005]; int lisan[20005],vv[20005]; void build(int idx,int s,int e) { node[idx].r=node[idx].g=node[idx].b=0; node[idx].cr=node[idx].cg=node[idx].cb=node[idx].crg=node[idx].crb=node[idx].cgb=node[idx].crgb=0; if(s!=e) { int mid=(s+e)>>1; build(idx<<1,s,mid); build(idx<<1|1,mid+1,e); } } void pop(int idx,int s,int e) { node[idx].cr=0; node[idx].cg=0; node[idx].cb=0; node[idx].crg=0; node[idx].crb=0; node[idx].cgb=0; node[idx].crgb=0; if(node[idx].r && node[idx].g && node[idx].b) { node[idx].crgb+=node[idx<<1].cr+node[idx<<1|1].cr; node[idx].crgb+=node[idx<<1].cg+node[idx<<1|1].cg; node[idx].crgb+=node[idx<<1].cb+node[idx<<1|1].cb; node[idx].crgb+=node[idx<<1].crg+node[idx<<1|1].crg; node[idx].crgb+=node[idx<<1].crb+node[idx<<1|1].crb; node[idx].crgb+=node[idx<<1].cgb+node[idx<<1|1].cgb; node[idx].crgb+=node[idx<<1].crgb+node[idx<<1|1].crgb; node[idx].crgb+=vv[e]-vv[s-1]-node[idx].cr-node[idx].cg-node[idx].cb-node[idx].crg-node[idx].crb-node[idx].cgb-node[idx].crgb; } if(!node[idx].r && node[idx].g && node[idx].b) { node[idx].crgb+=node[idx<<1].cr+node[idx<<1|1].cr; node[idx].cgb+=node[idx<<1].cg+node[idx<<1|1].cg; node[idx].cgb+=node[idx<<1].cb+node[idx<<1|1].cb; node[idx].crgb+=node[idx<<1].crg+node[idx<<1|1].crg; node[idx].crgb+=node[idx<<1].crb+node[idx<<1|1].crb; node[idx].cgb+=node[idx<<1].cgb+node[idx<<1|1].cgb; node[idx].crgb+=node[idx<<1].crgb+node[idx<<1|1].crgb; node[idx].cgb+=vv[e]-vv[s-1]-node[idx].cr-node[idx].cg-node[idx].cb-node[idx].crg-node[idx].crb-node[idx].cgb-node[idx].crgb; } if(node[idx].r && !node[idx].g && node[idx].b) { node[idx].crb+=node[idx<<1].cr+node[idx<<1|1].cr; node[idx].crgb+=node[idx<<1].cg+node[idx<<1|1].cg; node[idx].crb+=node[idx<<1].cb+node[idx<<1|1].cb; node[idx].crgb+=node[idx<<1].crg+node[idx<<1|1].crg; node[idx].crb+=node[idx<<1].crb+node[idx<<1|1].crb; node[idx].crgb+=node[idx<<1].cgb+node[idx<<1|1].cgb; node[idx].crgb+=node[idx<<1].crgb+node[idx<<1|1].crgb; node[idx].crb+=vv[e]-vv[s-1]-node[idx].cr-node[idx].cg-node[idx].cb-node[idx].crg-node[idx].crb-node[idx].cgb-node[idx].crgb; } if(node[idx].r && node[idx].g && !node[idx].b) { node[idx].crg+=node[idx<<1].cr+node[idx<<1|1].cr; node[idx].crg+=node[idx<<1].cg+node[idx<<1|1].cg; node[idx].crgb+=node[idx<<1].cb+node[idx<<1|1].cb; node[idx].crg+=node[idx<<1].crg+node[idx<<1|1].crg; node[idx].crgb+=node[idx<<1].crb+node[idx<<1|1].crb; node[idx].crgb+=node[idx<<1].cgb+node[idx<<1|1].cgb; node[idx].crgb+=node[idx<<1].crgb+node[idx<<1|1].crgb; node[idx].crg+=vv[e]-vv[s-1]-node[idx].cr-node[idx].cg-node[idx].cb-node[idx].crg-node[idx].crb-node[idx].cgb-node[idx].crgb; } if(node[idx].r && !node[idx].g && !node[idx].b) { node[idx].cr+=node[idx<<1].cr+node[idx<<1|1].cr; node[idx].crg+=node[idx<<1].cg+node[idx<<1|1].cg; node[idx].crb+=node[idx<<1].cb+node[idx<<1|1].cb; node[idx].crg+=node[idx<<1].crg+node[idx<<1|1].crg; node[idx].crb+=node[idx<<1].crb+node[idx<<1|1].crb; node[idx].crgb+=node[idx<<1].cgb+node[idx<<1|1].cgb; node[idx].crgb+=node[idx<<1].crgb+node[idx<<1|1].crgb; node[idx].cr+=vv[e]-vv[s-1]-node[idx].cr-node[idx].cg-node[idx].cb-node[idx].crg-node[idx].crb-node[idx].cgb-node[idx].crgb; } if(!node[idx].r && node[idx].g && !node[idx].b) { node[idx].crg+=node[idx<<1].cr+node[idx<<1|1].cr; node[idx].cg+=node[idx<<1].cg+node[idx<<1|1].cg; node[idx].cgb+=node[idx<<1].cb+node[idx<<1|1].cb; node[idx].crg+=node[idx<<1].crg+node[idx<<1|1].crg; node[idx].crgb+=node[idx<<1].crb+node[idx<<1|1].crb; node[idx].cgb+=node[idx<<1].cgb+node[idx<<1|1].cgb; node[idx].crgb+=node[idx<<1].crgb+node[idx<<1|1].crgb; node[idx].cg+=vv[e]-vv[s-1]-node[idx].cr-node[idx].cg-node[idx].cb-node[idx].crg-node[idx].crb-node[idx].cgb-node[idx].crgb; } if(!node[idx].r && !node[idx].g && node[idx].b) { node[idx].crb+=node[idx<<1].cr+node[idx<<1|1].cr; node[idx].cgb+=node[idx<<1].cg+node[idx<<1|1].cg; node[idx].cb+=node[idx<<1].cb+node[idx<<1|1].cb; node[idx].crgb+=node[idx<<1].crg+node[idx<<1|1].crg; node[idx].crb+=node[idx<<1].crb+node[idx<<1|1].crb; node[idx].cgb+=node[idx<<1].cgb+node[idx<<1|1].cgb; node[idx].crgb+=node[idx<<1].crgb+node[idx<<1|1].crgb; node[idx].cb+=vv[e]-vv[s-1]-node[idx].cr-node[idx].cg-node[idx].cb-node[idx].crg-node[idx].crb-node[idx].cgb-node[idx].crgb; } if(!node[idx].r && !node[idx].g && !node[idx].b)//别漏了这个 { node[idx].cr+=node[idx<<1].cr+node[idx<<1|1].cr; node[idx].cg+=node[idx<<1].cg+node[idx<<1|1].cg; node[idx].cb+=node[idx<<1].cb+node[idx<<1|1].cb; node[idx].crg+=node[idx<<1].crg+node[idx<<1|1].crg; node[idx].crb+=node[idx<<1].crb+node[idx<<1|1].crb; node[idx].cgb+=node[idx<<1].cgb+node[idx<<1|1].cgb; node[idx].crgb+=node[idx<<1].crgb+node[idx<<1|1].crgb; } } void update(int idx,int s,int e,int l,int r,int val,int flag) { if(l==s && r==e) { switch(val) { case 1:node[idx].r+=flag;break; case 2:node[idx].g+=flag;break; case 3:node[idx].b+=flag;break; } pop(idx,s,e); } else { int mid=(s+e)>>1; if(r<=mid) update(idx<<1,s,mid,l,r,val,flag); else if(l>mid) update(idx<<1|1,mid+1,e,l,r,val,flag); else { update(idx<<1,s,mid,l,mid,val,flag); update(idx<<1|1,mid+1,e,mid+1,r,val,flag); } pop(idx,s,e); } } int main() { int T,n,i,cnt,x1,y1,x2,y2,num,last,cases=1; long long ar,ag,ab,arg,arb,agb,argb; char s[5]; scanf("%d",&T); while(T--) { ii.clear(); ar=ag=ab=arg=arb=agb=argb=0; scanf("%d",&n); cnt=0; num=0; while(n--) { scanf("%s%d%d%d%d",s,&x1,&y1,&x2,&y2); if(!ii[y1]) lisan[num++]=y1,ii[y1]=1; if(!ii[y2]) lisan[num++]=y2,ii[y2]=1; line[cnt].pos=x1; line[cnt].l=y1; line[cnt].r=y2; line[cnt+1].pos=x2; line[cnt+1].l=y1; line[cnt+1].r=y2; if(s[0]=='R') { line[cnt].flag=1; line[cnt++].type=1; line[cnt].flag=-1; line[cnt++].type=1; } else if(s[0]=='G') { line[cnt].flag=1; line[cnt++].type=2; line[cnt].flag=-1; line[cnt++].type=2; } else { line[cnt].flag=1; line[cnt++].type=3; line[cnt].flag=-1; line[cnt++].type=3; } } sort(line,line+cnt); sort(lisan,lisan+num); for(i=0;i<num;i++) { ii[lisan[i]]=i+1; vv[i]=lisan[i]; } build(1,1,num); for(i=0;i<cnt;i++) { if(i) { ar+=(long long)(line[i].pos-last)*(node[1].cr); ag+=(long long)(line[i].pos-last)*(node[1].cg); ab+=(long long)(line[i].pos-last)*(node[1].cb); arg+=(long long)(line[i].pos-last)*(node[1].crg); arb+=(long long)(line[i].pos-last)*(node[1].crb); agb+=(long long)(line[i].pos-last)*(node[1].cgb); argb+=(long long)(line[i].pos-last)*(node[1].crgb); } update(1,1,num,ii[line[i].l],ii[line[i].r]-1,line[i].type,line[i].flag); last=line[i].pos; } printf("Case %d:\n",cases++); printf("%I64d\n",ar); printf("%I64d\n",ag); printf("%I64d\n",ab); printf("%I64d\n",arg); printf("%I64d\n",arb); printf("%I64d\n",agb); printf("%I64d\n",argb); } }
HDU-4419-Colourful Rectangle(线段树)
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