[THUPC2022 初赛] 画图 题解

题目链接

思路：

先把水平线段按y坐标排序，y坐标相同时把可以合并的合并，再把竖直线段按x坐标排序，x坐标相同时把可以合并的合并。

然后依次处理每个字母即可。先判断横竖线段数量是否够，如果不够，直接返回no。依次处理每个字母，枚举所有未被选择的线段，判断能否组成这个字母并且和其他字母不相交，如果所有字母都能组成，则返回yes，否则返回no。

  1 #include<cstdio>
  2 #include<algorithm>
  3 using namespace std;
  4 const int N=1e5+5;
  5 struct SHPI{
  6     int l,r,y;
  7 }sp[N],heng[N],T1,H4,U7,P10,P11,C14,C15;
  8 struct SHZH{
  9     int d,u,x;
 10 }sz[N],shu[N],T2,H3,H5,U6,U8,P9,P12,C13;
 11 int n,cntsp,cntsz,cnth,cnts;
 12 bool usedh[N],useds[N];
 13 int redd(){
 14   int xx=0,fl=1;char ch=getchar();
 15   while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
 16   if(ch=='-'){ch=getchar();fl=-1;}
 17   for(;ch>='0'&&ch<='9';ch=getchar())xx=(xx<<3)+(xx<<1)+ch-'0';
 18   return xx*fl;
 19 }
 20 int maxx(int x,int y){
 21     return x>y?x:y;
 22 }
 23 int minn(int x,int y){
 24     return x<y?x:y;
 25 }
 26 bool cmpsp(SHPI p,SHPI q){
 27     if(p.y!=q.y)return p.y<q.y;
 28     return p.l<q.l;
 29 }
 30 bool cmpsz(SHZH p,SHZH q){
 31     if(p.x!=q.x)return p.x<q.x;
 32     return p.d<q.d;
 33 }
 34 bool xj(SHPI p,SHZH q){                 //判断是否相交
 35     if(p.l<=q.x&&q.x<=p.r&&q.d<=p.y&&p.y<=q.u)return 1;
 36     return 0;
 37 }
 38 bool workt(){                          //T
 39     for(int i=1;i<=cnth;i++){
 40         for(int j=1;j<=cnts;j++){
 41             T1=heng[i];T2=shu[j];usedh[i]=1;useds[j]=1;
 42             if(T2.d<T1.y&&T1.y==T2.u&&T1.l<T2.x&&T2.x<T1.r)return 1;
 43             usedh[i]=0;useds[j]=0;
 44         }
 45     }
 46     return 0;
 47 }
 48 bool workh(){                          //H
 49     for(int i=1;i<=cnts;i++)if(!useds[i]){
 50         H3=shu[i];
 51         if(xj(T1,H3))continue;
 52         useds[i]=1;
 53         for(int j=1;j<=cnth;j++)if(!usedh[j]){
 54             H4=heng[j];
 55             if(xj(H4,T2))continue;
 56             usedh[j]=1;
 57             for(int k=1;k<=cnts;k++)if(!useds[k]){
 58                 H5=shu[k];
 59                 if(xj(T1,H5))continue;
 60                 useds[k]=1;
 61                 if(H3.d==H5.d&&H5.d<H4.y&&H4.y<H3.u&&H3.u==H5.u&&H3.x==H4.l&&H4.l<H4.r&&H4.r==H5.x)return 1;
 62                 useds[k]=0;
 63             }
 64             usedh[j]=0;
 65         }
 66         useds[i]=0;
 67     }
 68     return 0;
 69 }
 70 bool worku(){                          //U
 71     for(int i=1;i<=cnts;i++)if(!useds[i]){
 72         U6=shu[i];
 73         if(xj(T1,U6))continue;
 74         if(xj(H4,U6))continue;
 75         useds[i]=1;
 76         for(int j=1;j<=cnth;j++)if(!usedh[j]){
 77             U7=heng[j];
 78             if(xj(U7,T2))continue;
 79             if(xj(U7,H3))continue;
 80             if(xj(U7,H5))continue;
 81             usedh[j]=1;
 82             for(int k=1;k<=cnts;k++)if(!useds[k]){
 83                 U8=shu[k];
 84                 if(xj(T1,U8))continue;
 85                 if(xj(H4,U8))continue;
 86                 useds[k]=1;
 87                 if(U6.d==U8.d&&U8.d==U7.y&&U7.y<U6.u&&U6.u==U8.u&&U6.x==U7.l&&U7.l<U7.r&&U7.r==U8.x)return 1;
 88                 useds[k]=0;
 89             }
 90             usedh[j]=0;
 91         }
 92         useds[i]=0;
 93     }
 94     return 0;
 95 }
 96 bool workp(){                          //P
 97     for(int i=1;i<=cnts;i++)if(!useds[i]){
 98         P9=shu[i];
 99         if(xj(T1,P9))continue;
100         if(xj(H4,P9))continue;
101         if(xj(U7,P9))continue;
102         useds[i]=1;
103         for(int j=1;j<=cnth;j++)if(!usedh[j]){
104             P10=heng[j];
105             if(xj(P10,T2))continue;
106             if(xj(P10,H3))continue;
107             if(xj(P10,H5))continue;
108             if(xj(P10,U6))continue;
109             if(xj(P10,U8))continue;
110             usedh[j]=1;
111             for(int k=1;k<=cnth;k++)if(!usedh[k]){
112                 P11=heng[k];
113                 if(xj(P11,T2))continue;
114                 if(xj(P11,H3))continue;
115                 if(xj(P11,H5))continue;
116                 if(xj(P11,U6))continue;
117                 if(xj(P11,U8))continue;
118                 usedh[k]=1;
119                 for(int l=1;l<=cnts;l++)if(!useds[l]){
120                     P12=shu[l];
121                     if(xj(T1,P12))continue;
122                     if(xj(H4,P12))continue;
123                     if(xj(U7,P12))continue;
124                     useds[l]=1;
125                     if(P9.d<P11.y&&P11.y==P12.d&&P12.d<P9.u&&P9.u==P10.y&&P10.y==P12.u&&P9.x==P10.l&&P10.l==P11.l&&P11.l<P10.r&&P10.r==P11.r&&P11.r==P12.x)return 1;
126                     useds[l]=0;
127                 }
128                 usedh[k]=0;
129             }
130             usedh[j]=0;
131         }
132         useds[i]=0;
133     }
134     return 0;
135 }
136 bool workc(){                          //C
137     for(int i=1;i<=cnts;i++)if(!useds[i]){
138         C13=shu[i];
139         if(xj(T1,C13))continue;
140         if(xj(H4,C13))continue;
141         if(xj(U7,C13))continue;
142         if(xj(P10,C13))continue;
143         if(xj(P11,C13))continue;
144         useds[i]=1;
145         for(int j=1;j<=cnth;j++)if(!usedh[j]){
146             C14=heng[j];
147             if(xj(C14,T2))continue;
148             if(xj(C14,H3))continue;
149             if(xj(C14,H5))continue;
150             if(xj(C14,U6))continue;
151             if(xj(C14,U8))continue;
152             if(xj(C14,P9))continue;
153             if(xj(C14,P12))continue;
154             usedh[j]=1;
155             for(int k=1;k<=cnth;k++)if(!usedh[k]){
156                 C15=heng[k];
157                 if(xj(C15,T2))continue;
158                 if(xj(C15,H3))continue;
159                 if(xj(C15,H5))continue;
160                 if(xj(C15,U6))continue;
161                 if(xj(C15,U8))continue;
162                 if(xj(C15,P9))continue;
163                 if(xj(C15,P12))continue;
164                 usedh[k]=1;
165                 if(C13.d==C15.y&&C15.y<C13.u&&C13.u==C14.y&&C13.x==C14.l&&C14.l==C15.l&&C15.l<C14.r&&C14.r==C15.r)return 1;
166                 usedh[k]=0;
167             }
168             usedh[j]=0;
169         }
170         useds[i]=0;
171     }
172     return 0;
173 }
174 bool work(){
175     sort(sp+1,sp+cntsp+1,cmpsp);
176     sort(sz+1,sz+cntsz+1,cmpsz);
177     for(int i=1;i<=cntsp;i++){
178         if(!cnth||sp[i].y!=heng[cnth].y){
179             cnth++;heng[cnth]=sp[i];continue;
180         }
181         if(sp[i].l>heng[cnth].r){
182             cnth++;heng[cnth]=sp[i];continue;
183         }
184         heng[cnth].r=maxx(heng[cnth].r,sp[i].r);
185     }
186     for(int i=1;i<=cntsz;i++){
187         if(!cnts||sz[i].x!=shu[cnts].x){
188             cnts++;shu[cnts]=sz[i];continue;
189         }
190         if(sz[i].d>shu[cnts].u){
191             cnts++;shu[cnts]=sz[i];continue;
192         }
193         shu[cnts].u=maxx(shu[cnts].u,sz[i].u);
194     }
195     if(cnth!=7)return 0;
196     if(cnts!=8)return 0;
197     if(!workt())return 0;
198     if(!workh())return 0;
199     if(!worku())return 0;
200     if(!workp())return 0;
201     if(!workc())return 0;
202     return 1;
203 }
204 int main(){
205     // freopen("1.in","r",stdin);
206     // freopen("1.out","w",stdout);
207     int opt;
208     n=redd();
209     for(int i=1;i<=n;i++){
210         opt=redd();
211         if(!opt){
212             cntsp++;
213             sp[cntsp].l=redd();sp[cntsp].r=redd();sp[cntsp].y=redd();
214         }
215         else{
216             cntsz++;
217             sz[cntsz].d=redd();sz[cntsz].u=redd();sz[cntsz].x=redd();
218         }
219     }
220     if(work())printf("Yes\n");
221     else printf("No\n");
222   return 0;
223 }

 

posted @ 2022-03-22 19:23  le0632  阅读(139)  评论(1)    收藏  举报