狗狗40题~ (Volume C)
A - Triangles
记忆化搜索呗。搜索以某三角形为顶的最大面积，注意边界情况。
1 #include <stdio.h>
2 #include <cstring>
3 #include <cstdlib>
4 #include <algorithm>
5 #include <cmath>
6
7 using namespace std; 8 #define lson o<<1
9 #define rson o<<1|1
10 #define max(a,b) (a)>(b)?(a):(b)
11 #define min(a,b) (a)<(b)?(a):(b)
12 #define INF 200000000
13
14 typedef long long ll; 15 int dp[110][220]; 16 char g[110][220]; 17 int main(){ 18 int n,h,cs=1; 19 while(scanf("%d",&n) && n){ 20 for(int i=0;i<n;i++)scanf("%s",g[i]); 21 memset(dp,0,sizeof dp); 22
23 printf("Triangle #%d\n",cs++); 24 if(n==1){ 25 printf("The largest triangle area is %d.\n\n",g[0][0]=='#'?0:1); 26 continue; 27 } 28
29 h=0; 30 for(int j=0;j<2*n-1;j++)if(g[0][j]!='#')dp[0][j]=h=1; 31 for(int i=1;i<n;i++){ 32 int j; 33 for(j=0;j<2*(n-i);j+=2)if(g[i][j]!='#'){ 34 int f=i-1,p=j+1; 35 while(f>=0&&g[f][p]!='#'&&g[f][p+1]!='#')f--,p+=2; 36 dp[i][j]=min(1+dp[i-1][j],i-f); 37 h=max(h,dp[i][j]); 38 } 39 } 40 if(g[n-2][1]!='#'){dp[n-2][1]=1;h=max(h,1);} 41 for(int i=n-3;i>=0;i--){ 42 int j; 43 for(j=1;j<2*(n-i)-1;j+=2)if(g[i][j]!='#'){ 44 if(j<2)dp[i][j]=1; 45 else{ 46 int f=i+1,p=j-1; 47 while(f<n && j<2*(n-f)-1 && g[f][p]!='#' && g[f][p+1]!='#')f++; 48 dp[i][j]=min(1+dp[i+1][j-2],f-i); 49 } 50 h=max(h,dp[i][j]); 51 } 52 } 53 printf("The largest triangle area is %d.\n\n",h*h); 54 } 55 return 0; 56 }
B - Domino Effect
基本就是最短路问题，权值为正，最后要用 t=(dis[i]+dis[j]+g[i][j])/2.0 计算边上最后一个多米诺倒下的时间。
1 #include <stdio.h>
2 #include <cstring>
3 #include <cstdlib>
4 #include <algorithm>
5 #include <cmath>
6 #include <queue>
7
8 using namespace std; 9 #define lson o<<1
10 #define rson o<<1|1
11 #define max(a,b) (a)>(b)?(a):(b)
12 #define min(a,b) (a)<(b)?(a):(b)
13 #define INF 20000000000LL
14
15 typedef long long ll; 16 int g[505][505],n,m,vis[505]; 17 ll d[505]; 18 double t;int x,y; 19 void dijkstra(){ 20 for(int i=2;i<=n;i++)d[i]=INF; 21 d[1]=0; 22 for(int i=0;i<n;i++){ 23 int u; 24 ll tm=INF; 25 for(int j=1;j<=n;j++)if(!vis[j]&&d[j]<tm){ 26 u=j;tm=d[j]; 27 } 28 if(tm>=INF)break; 29 vis[u]=1; 30 for(int j=1;j<=n;j++)if(g[u][j]!=-1&&!vis[j]) 31 d[j]=min(d[j],g[u][j]+d[u]); 32 } 33 for(int i=1;i<=n;i++){ 34 if(t<d[i]){x=i;y=0;t=d[i];} 35 for(int j=1;j<i;j++)if(g[i][j]!=-1){ 36 double temp=(g[i][j]+d[i]+d[j])/2.0; 37 if(t<temp){x=j;y=i;t=temp;} 38 } 39
40 } 41 } 42 int main(){ 43 int a,b,l,cs=1; 44 while(scanf("%d%d",&n,&m) && n){ 45 memset(g,-1,sizeof g); 46 memset(vis,0,sizeof vis); 47 t=0;x=1;y=0; 48 for(int i=0;i<m;i++){ 49 scanf("%d%d%d",&a,&b,&l); 50 g[a][b]=g[b][a]=l; 51 } 52 dijkstra(); 53 printf("System #%d\n",cs++); 54 if(!y){ 55 printf("The last domino falls after %.1lf seconds, at key domino %d.\n\n",t,x); 56 }else{ 57 printf("The last domino falls after %.1lf seconds, between key dominoes %d and %d.\n\n",t,x,y); 58 } 59 } 60 return 0; 61 }
C - Pendulum
代码还比较简短，注意转的时候不能高于x轴
1 #include <stdio.h> 2 #include <cstring> 3 #include <cstdlib> 4 #include <algorithm> 5 #include <cmath> 6 7 using namespace std; 8 #define lson o<<1 9 #define rson o<<1|1 10 #define max(a,b) (a)>(b)?(a):(b) 11 #define min(a,b) (a)<(b)?(a):(b) 12 #define INF 2000000000 13 #define eps 1e-6 14 15 typedef long long ll; 16 const double pi=acos(-1); 17 struct point{ 18 int x,y; 19 point() {} 20 point(int x,int y): x(x),y(y) {} 21 point operator + (const point a){return point(x+a.x,y+a.y);} 22 point operator - (const point a){return point(x-a.x,y-a.y);} 23 }p[505]; 24 int n; 25 int dcmp(double x){ 26 if(x>eps)return 1; 27 else if(x<-eps)return -1; 28 return 0; 29 } 30 double dist(point c){ 31 return sqrt((double)c.x*c.x+(double)c.y*c.y); 32 } 33 double ans; 34 void swing(int o,double l,double ang){ 35 double a0=2*pi,dm=0.0; 36 int next=-1; 37 double amax=2*pi; 38 if(dcmp(p[o].y-l)<0)amax=pi+asin(p[o].y/l)-ang; 39 for(int i=1;i<n;i++)if(i!=o){ 40 double d=dist(p[i]-p[o]); 41 if(dcmp(d-l)>=0)continue; 42 double a=atan2(p[i].y-p[o].y,p[i].x-p[o].x)-ang; 43 while(a<0)a+=2*pi; 44 if(dcmp(a-amax)>=0)continue; 45 if(dcmp(a-a0)<0||(dcmp(a-a0)==0&&d>dm)){ 46 dm=d,a0=a; 47 next=i; 48 } 49 } 50 if(next!=-1){ 51 ans+=a0*l; 52 double a=a0+ang; 53 if(a>=2*pi)a-=2*pi; 54 swing(next,l-dm,a); 55 } 56 else if(dcmp(p[o].y-l)<0)ans+=amax*l; 57 else ans=pi*l; 58 } 59 60 int main(){ 61 int cs=1; 62 double r; 63 while(scanf("%d%lf",&n,&r) && dcmp(r)){ 64 n++; 65 p[0]=point(0,0); 66 for(int i=1;i<n;i++){ 67 scanf("%d%d",&p[i].x,&p[i].y); 68 p[i].x=-p[i].x,p[i].y=-p[i].y; 69 } 70 ans=0.0; 71 swing(0,r,0.0); 72 printf("Pendulum #%d\n",cs++); 73 printf("Length of periodic orbit = %.2lf\n\n",ans*2); 74 } 75 return 0; 76 }
D - The New Villa
这种题应该一看数据规模就知道是bfs可以水过的。
1 #include <stdio.h>
2 #include <cstdlib>
3 #include <cstring>
4 #include <algorithm>
5
6 #define STA 30000
7 int q[STA],d[STA],fa[STA],vis[STA],ans[STA]; 8 int to[15][15],sw[15][15]; 9 int n,k,s; 10 int bfs(int st){ 11 int l,r; 12 l=r=0; 13 int now=(1<<st)*10+st-1; 14 q[r++]=now; 15 fa[now]=-1; 16 while(l<r){ 17 now=q[l++]; 18 int room=now%10+1,sta=now/10; 19 if(room==n&&sta==(1<<n))return 1; 20 int next; 21 for(int i=1;i<=n;i++)if(!vis[now-room+i] && to[room][i] && sta&(1<<i)){ 22 next=now-room+i; 23 vis[next]=1; 24 d[next]=d[now]+1; 25 fa[next]=now; 26 q[r++]=next; 27 } 28 for(int i=1;i<=n;i++)if(i!=room && sw[room][i]){ 29 if(!(sta&(1<<i)) && !vis[(sta|(1<<i))*10+room-1]){ 30 next=(sta|(1<<i))*10+room-1; 31 vis[next]=1; 32 d[next]=d[now]+1; 33 fa[next]=now; 34 q[r++]=next; 35 } 36 else if(sta&(1<<i) && !vis[(sta^(1<<i))*10+room-1]){ 37 next=(sta^(1<<i))*10+room-1; 38 vis[next]=1; 39 d[next]=d[now]+1; 40 fa[next]=now; 41 q[r++]=next; 42 } 43 } 44 } 45 return 0; 46
47 } 48 int main(){ 49 //freopen("input.in","r",stdin);freopen("output.out","w",stdout);
50 int cs=1; 51 while(scanf("%d%d%d",&n,&k,&s) &&n){ 52 int x,y; 53 memset(to,0,sizeof to); 54 memset(sw,0,sizeof sw); 55 memset(vis,0,sizeof vis); 56 memset(d,0,sizeof d); 57 while(k--){ 58 scanf("%d%d",&x,&y); 59 to[x][y]=to[y][x]=1; 60 } 61 while(s--){ 62 scanf("%d%d",&x,&y); 63 sw[x][y]=1; 64 } 65 printf("Villa #%d\n",cs++); 66 if(bfs(1)){ 67 int u=(1<<n)*10+n-1,dis; 68 printf("The problem can be solved in %d steps:\n",dis=d[u]); 69 for(int i=0;i<dis;i++){ 70 ans[i]=u; 71 u=fa[u]; 72 } 73 int lastroom=1,sta0=1<<1,room,sta; 74 for(int i=dis-1;i>=0;i--){ 75 room=ans[i]%10+1;sta=ans[i]/10; 76 if(room!=lastroom)printf("- Move to room %d.\n",room); 77 else{ 78 for(int j=1;j<=n;j++)if((sta0&(1<<j))!=(sta&(1<<j))){ 79 if(sta0&(1<<j))printf("- Switch off light in room %d.\n",j); 80 else printf("- Switch on light in room %d.\n",j); 81 break; 82 } 83 } 84 lastroom=room;sta0=sta; 85 } 86 printf("\n"); 87
88 }else printf("The problem cannot be solved.\n\n"); 89
90 } 91 return 0; 92 }
E - Parallelepiped Walk
巧妙地转化成坐标旋转的问题，有几个简化代码的技巧
1 #include <stdio.h> 2 #include <cstring> 3 #include <cstdlib> 4 #include <algorithm> 5 #include <cmath> 6 7 using namespace std; 8 #define lson o<<1 9 #define rson o<<1|1 10 #define max(a,b) (a)>(b)?(a):(b) 11 #define min(a,b) (a)<(b)?(a):(b) 12 #define INF 2000000000 13 14 typedef long long ll; 15 int xx,yy,zz; 16 ll ans; 17 ll dist(int x,int y){ 18 return (ll)x*x+(ll)y*y; 19 } 20 void go(int dx,int dy,int x,int y,int z,int x1,int x2,int y1,int y2,int h){ 21 if(z==0){ 22 ans=min(ans,dist(x-xx,y-yy)); 23 return; 24 } 25 if(dx>=0&&dx<2)go(dx+1,dy,x2+z,y,x2-x,x2,x2+h,y1,y2,x2-x1); 26 if(dx<=0&&dx>-2)go(dx-1,dy,x1-z,y,x-x1,x1-h,x1,y1,y2,x2-x1); 27 if(dy>=0&&dy<2)go(dx,dy+1,x,y2+z,y2-y,x1,x2,y2,y2+h,y2-y1); 28 if(dy<=0&&dy>-2)go(dx,dy-1,x,y1-z,y-y1,x1,x2,y1-h,y1,y2-y1); 29 } 30 int main(){ 31 int x0,y0,z0,x,y,z; 32 while(~scanf("%d%d%d%d%d%d%d%d%d",&x0,&y0,&z0,&xx,&yy,&zz,&x,&y,&z)){ 33 if(xx==0||xx==x0){ 34 swap(xx,zz); 35 swap(x0,z0); 36 swap(x,z); 37 } 38 else if(yy==0||yy==y0){ 39 swap(yy,zz); 40 swap(y0,z0); 41 swap(y,z); 42 } 43 if(zz==z0){ 44 zz=z0-zz; 45 z=z0-z; 46 } 47 ans=INF; 48 go(0,0,x,y,z,0,x0,0,y0,z0); 49 printf("%lld\n",ans); 50 } 51 return 0; 52 }
F - Decoding Morse Sequences
我的做法是把所有单词的Morse码编进Trie树，在词尾标记访问次数，然后记忆化搜索。
1 #include <stdio.h>
2 #include <cstring>
3 #include <cstdlib>
4 #include <algorithm>
5 #include <cmath>
6
7 using namespace std; 8 #define lson o<<1
9 #define rson o<<1|1
10 #define max(a,b) (a)>(b)?(a):(b)
11 #define min(a,b) (a)<(b)?(a):(b)
12 #define INF 200000000
13
14 typedef long long ll; 15 int ch[1000010][3],sz; 16 char morse[26][10] 17 ={".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..", 18 "--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."}; 19 char s[10100],word[25],mod[100]; 20 ll dp[10100]; 21 int t[1000010]; 22
23 void insert(char *s){ 24 char *p=s; 25 int u=0; 26 for(;*p!='\0';p++){ 27 int temp; 28 if(*p=='.')temp=1; 29 else temp=2; 30 if(ch[u][temp]==-1)ch[u][temp]=++sz; 31 u=ch[u][temp]; 32 } 33 if(ch[u][0]==-1)ch[u][0]=++sz; 34 u=ch[u][0]; 35 t[u]++; 36 } 37 ll dfs(int now){ 38 if(dp[now]>=0)return dp[now]; 39 ll &ans=dp[now]; 40 ans=0; 41 int u=0,flag=1; 42 for(int i=now;s[i];i++){ 43 int temp; 44 if(s[i]=='.')temp=1; 45 else temp=2; 46 if(ch[u][temp]==-1)return ans; 47 u=ch[u][temp]; 48 if(ch[u][0]!=-1)ans+=t[ch[u][0]]*dfs(i+1); 49 } 50 return ans; 51 } 52 int main(){ 53 //freopen("r.in","r",stdin);freopen("r.out","w",stdout);
54 int T; 55 scanf("%d",&T); 56 while(T--){ 57 scanf("%s",s); 58 int m; 59 scanf("%d",&m); 60 memset(ch,-1,sizeof ch); 61 memset(t,0,sizeof t); 62 sz=0; 63 for(int i=0;i<m;i++){ 64 mod[0]='\0'; 65 scanf("%s",word); 66 for(int j=0;word[j];j++) 67 strcat(mod,morse[word[j]-'A']); 68 insert(mod); 69 } 70 memset(dp,-1,sizeof dp); 71 dp[strlen(s)]=1; 72 dfs(0); 73 printf("%lld\n",dp[0]); 74 } 75 return 0; 76 }
G - Fill the Cisterns!
二分答案，精度0.0005.
1 #include <stdio.h>
2 #include <cstring>
3 #include <cstdlib>
4 #include <algorithm>
5 #include <cmath>
6
7 using namespace std; 8 #define lson o<<1
9 #define rson o<<1|1
10 #define max(a,b) (a)>(b)?(a):(b)
11 #define min(a,b) (a)<(b)?(a):(b)
12 #define INF 2000000
13 #define eps 5e-4
14
15 typedef long long ll; 16
17 struct cistern{ 18 double b,h,s; 19 }c[50010]; 20 double v,hmax,hmin; 21 int n; 22
23 int cmp(cistern a,cistern b){return a.b<b.b;} 24
25 int cal(double dh){ 26 double vx=0; 27 for(int i=0;i<n&&c[i].b<dh;i++){ 28 double t=min(c[i].b+c[i].h,dh); 29 vx+=(t-c[i].b)*c[i].s; 30 } 31 if(vx<v)return -1; 32 if(vx>v)return 1; 33 return 0; 34 } 35 double search(double low,double hi){ 36 double mid; 37 while(hi-low>eps){ 38 mid=(hi+low)/2.0; 39 if(cal(mid)>=0)hi=mid; 40 else low=mid; 41 } 42 return hi; 43 } 44 int main(){ 45 //freopen("r.in","r",stdin);freopen("r.out","w",stdout);
46 int t; 47 scanf("%d",&t); 48 while(t--){ 49 scanf("%d",&n); 50 double w,h,vmax; 51 hmax=0;hmin=INF; 52 vmax=0; 53 for(int i=0;i<n;i++){ 54 scanf("%lf%lf%lf%lf",&c[i].b,&c[i].h,&w,&h); 55 c[i].s=w*h; 56 hmax=max(hmax,c[i].h+c[i].b); 57 hmin=min(hmin,c[i].b); 58 vmax+=c[i].s*c[i].h; 59 } 60 scanf("%lf",&v); 61 if(vmax<v){ 62 printf("OVERFLOW\n"); 63 continue; 64 } 65
66 sort(c,c+n,cmp); 67 double dh=search(hmin,hmax); 68 printf("%.2lf\n",dh); 69
70 } 71 return 0; 72 }
H - Horizontally Visible Segments
线段树染色问题……每次询问一次更新一次，中间hash判重然后加边，最后暴力搜索得到 Triangle 个数。
有个细节是x,y在[0,8000]，0<=n<=8000,建立的图应该是稀疏图，用vector存，所以一定要判重的！
1 #include <stdio.h>
2 #include <cstdlib>
3 #include <cstring>
4 #include <algorithm>
5 #include <vector>
6
7 #define rson o<<1|1
8 #define lson o<<1
9 #define H 16000
10 #define MOD 1403641
11
12 using namespace std; 13
14 vector <int> g[8010]; 15 int col[16000<<2],y1,y2; 16
17 struct segment{ 18 int y1,y2,x; 19 }seg[8010]; 20 struct Hash{ 21 int va; 22 Hash *next; 23 Hash(){va=-1;next=NULL;} 24 }hash[MOD]; 25 void init(){ 26 for(int i=0;i<MOD;i++){hash[i].va=-1;hash[i].next=NULL;} 27 } 28 int exist(int u,int v){ 29 int x=u*10000+v; 30 if(hash[x%MOD].va==-1){hash[x%MOD].va=x;return 0;} 31 else{ 32 Hash *u=&hash[x%MOD]; 33 while(u->next!=NULL){ 34 if(u->va==x)return 1; 35 u=u->next; 36 } 37 if(u->va==x)return 1; 38 u->next=new Hash; 39 u->next->va=x;u->next->next=NULL; 40 return 0; 41 } 42 } 43 int cmp(struct segment a,struct segment b){return a.x<b.x;} 44 void pushdown(int o,int l,int r){ 45 if(col[o]!=-1){ 46 col[rson]=col[lson]=col[o]; 47 } 48 } 49 void upd(int o,int l,int r,int u){ 50 if(y1<=l&&y2>=r){ 51 col[o]=u; 52 }else if(y2>=l&&y1<=r){ 53 int mid=(l+r)>>1; 54 pushdown(o,l,r); 55 col[o]=-1; 56 if(y1<=mid)upd(lson,l,mid,u); 57 if(y2>mid)upd(rson,mid+1,r,u); 58 } 59 } 60 void que(int o,int l,int r,int v){ 61 int mid=(l+r)/2; 62 if(y1<=l&&y2>=r){ 63 if(col[o]==-1){ 64 que(lson,l,mid,v); 65 que(rson,mid+1,r,v); 66 }else if(col[o]&&!exist(v,col[o])){ 67 g[v].push_back(col[o]); 68 } 69 }else if(y2>=l&&y1<=r){ 70 pushdown(o,l,r); 71 if(y1<=mid)que(lson,l,mid,v); 72 if(y2>mid)que(rson,mid+1,r,v); 73 } 74 } 75 int main(){ 76 int T,n; 77 scanf("%d",&T); 78 while(T--){ 79 scanf("%d",&n); 80 for(int i=1;i<=n;i++)scanf("%d%d%d",&seg[i].y1,&seg[i].y2,&seg[i].x); 81 sort(seg+1,seg+n+1,cmp); 82
83 for(int i=1;i<=n;i++)g[i].clear(); 84 memset(col,0,sizeof col); 85 init(); 86 for(int i=1;i<=n;i++){ 87 y1=seg[i].y1*2;y2=seg[i].y2*2; 88 if(i>1)que(1,0,H,i); 89 upd(1,0,H,i); 90 } 91 int cnt=0; 92 for(int i=3;i<=n;i++)if(g[i].size()>1) 93 for(int j=0;j<g[i].size();j++){ 94 int u=g[i][j]; 95 for(int k=0;k<g[u].size();k++) 96 for(int q=0;q<g[i].size();q++)if(g[i][q]==g[u][k])cnt++; 97 } 98
99 printf("%d\n",cnt); 100 } 101 return 0; 102 }
I - 2D Nim
按照每个点向四个方向走的最大步数来判断，哈希查找
1 #include <stdio.h> 2 #include <cstring> 3 #include <cstdlib> 4 #include <algorithm> 5 #include <cmath> 6 using namespace std; 7 #define lson o<<1 8 #define rson o<<1|1 9 #define max(a,b) (a)>(b)?(a):(b) 10 #define min(a,b) (a)<(b)?(a):(b) 11 #define INF 2000000000 12 #define M 240 13 14 typedef long long ll; 15 int g[120][120]; 16 int n,m; 17 struct hash_table{ 18 int v[4]; 19 bool vis; 20 hash_table* next; 21 hash_table(){vis=0;next=NULL;} 22 void init(){ 23 vis=0; 24 next=NULL; 25 } 26 void insert(int *t){ 27 if(!vis){ 28 for(int i=0;i<4;i++)v[i]=t[i]; 29 vis=1; 30 } 31 else{ 32 hash_table *u=new hash_table; 33 u->next=next; 34 next=u; 35 u->vis=1; 36 for(int i=0;i<4;i++)u->v[i]=t[i]; 37 } 38 } 39 int search(int *t){ 40 if(vis){ 41 bool flag=1; 42 for(int i=0;i<4;i++)if(v[i]!=t[i])flag=0; 43 if(flag){vis=0;return 1;} 44 } 45 hash_table* now=next; 46 while(now!=NULL){ 47 if(now->vis){ 48 bool flag=1; 49 for(int i=0;i<4;i++)if(now->v[i]!=t[i])flag=0; 50 if(flag){now->vis=0;return 1;} 51 } 52 now=now->next; 53 } 54 return 0; 55 } 56 }hash[M]; 57 int main(){ 58 // freopen("r.in","r",stdin);freopen("r.out","w",stdout); 59 int t; 60 scanf("%d",&t); 61 while(t--){ 62 int k; 63 scanf("%d%d%d",&n,&m,&k); 64 memset(g,0,sizeof g); 65 for(int i=0;i<M;i++)hash[i].init(); 66 int x,y; 67 for(int i=0;i<k;i++){ 68 scanf("%d%d",&x,&y); 69 g[x][y]=1; 70 } 71 int tm[4]; 72 for(int i=0;i<n;i++) 73 for(int j=0;j<m;j++)if(g[i][j]){ 74 int tot=0; 75 memset(tm,0,sizeof tm); 76 for(int x=i-1;x>=0&&g[x][j];x--)tm[0]++,tot++; 77 for(int x=i+1;x<n&&g[x][j];x++)tm[1]++,tot++; 78 for(int y=j-1;y>=0&&g[i][y];y--)tm[2]++,tot++; 79 for(int y=j+1;y<m&&g[i][y];y++)tm[3]++,tot++; 80 sort(tm,tm+4); 81 hash[tot].insert(tm); 82 } 83 memset(g,0,sizeof g); 84 for(int i=0;i<k;i++){ 85 scanf("%d%d",&x,&y); 86 g[x][y]=1; 87 } 88 bool flag=1; 89 for(int i=0;i<n;i++) 90 for(int j=0;j<m;j++)if(g[i][j]){ 91 int tot=0; 92 memset(tm,0,sizeof tm); 93 for(int x=i-1;x>=0&&g[x][j];x--)tm[0]++,tot++; 94 for(int x=i+1;x<n&&g[x][j];x++)tm[1]++,tot++; 95 for(int y=j-1;y>=0&&g[i][y];y--)tm[2]++,tot++; 96 for(int y=j+1;y<m&&g[i][y];y++)tm[3]++,tot++; 97 sort(tm,tm+4); 98 if(!hash[tot].search(tm)){flag=0;break;} 99 } 100 101 if(flag)printf("YES\n"); 102 else printf("NO\n"); 103 } 104 return 0; 105 }
J - (Your)((Term)((Project)))
细节题了，递归做最保险。
1.skip空格。
2.加号（或无符号默认为加号）后面对应括号去掉。
3.重复的空格去掉。
1 #include <stdio.h>
2 #include <cstring>
3 #include <cstdlib>
4 #include <algorithm>
5 #include <cmath>
6
7 using namespace std; 8 #define lson o<<1
9 #define rson o<<1|1
10 #define max(a,b) (a)>(b)?(a):(b)
11 #define min(a,b) (a)<(b)?(a):(b)
12 #define INF 200000000
13 #define SKIP(p) while(*p==' ')++p;
14
15 typedef long long ll; 16 char s[300],q[300]; 17 int res; 18 void solve(char *s,char *q){ 19 char *p; 20 int sgn=1; 21 for(p=s;p<=q;p++){ 22 if(*p=='+')sgn=1; 23 else if(*p=='-')sgn=-1; 24 else if(*p=='('){ 25 char *t; 26 int cnt=1; 27 for(t=p+1;t<=q;t++){ 28 if(*t=='(')cnt++; 29 else if(*t==')')cnt--; 30 if(!cnt)break; 31 } 32 if(sgn>0)*p=*t=' '; 33 if(t-p>2)solve(p+1,t-1); 34 } 35 } 36 } 37 void deal(char *p,int n){ 38 for(int i=0;i<n;i++)if(p[i]<='Z'&&p[i]>='A'){ 39 int l=i-1,r=i+1; 40 while(l>=0&&r<n){ 41 if(p[l]!='('||p[r]!=')')break; 42 p[l]=p[r]=' '; 43 l--;r++; 44 } 45 } 46 solve(p,p+n-1); 47 } 48 int main(){ 49 //freopen("r.in","r",stdin);freopen("r.out","w",stdout);
50 int t; 51 scanf("%d",&t); 52 gets(s); 53 while(t--){ 54 gets(s); 55 char *p=s; 56 res=0; 57 for(;*p!='\0';p++){ 58 SKIP(p); 59 q[res++]=*p; 60 } 61 deal(q,res); 62 for(int i=0;i<res;i++)if(q[i]!=' ') 63 printf("%c",q[i]); 64 printf("\n"); 65 } 66 return 0; 67 }