UVaOJ 112道题目-排序
1、110401/10041 Vito’s Family (Vito 家族)
距离最小的点必定是中位数,必定出现在输入的点之间
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#include<stdio.h> #include<string.h> #include<algorithm> #include<math.h> #include<ctype.h> using namespace std; int s[505]; int dis[30005]; const int INF=1000000000; int main() { int T,r; scanf("%d",&T); while(T--) { scanf("%d",&r); int j,i,sum=0; memset(dis,0,sizeof(dis)); for(i=0;i<r;i++) { scanf("%d",&s[i]); } for(i=0;i<r;i++) { int x=s[i]; for(j=0;j<r;j++) { dis[x]+=abs(s[j]-x); } } int d=INF; for(i=0;i<r;i++) if(dis[s[i]]<d)d=dis[s[i]]; printf("%d\n",d); } return 0; }
2、110402/120 Stacks of Flapjacks (煎饼堆)
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#include<stdio.h> #include<string.h> #include<algorithm> #include<math.h> #include<ctype.h> using namespace std; int p[35]; int s[35]; char cmd[100]; int k=0; void read() { int len=strlen(cmd),i,tot=0; k=0; for(i=0;i<=len;i++) { if(i<len&&cmd[i]!=' ') { tot=tot*10+cmd[i]-'0'; } else { if(tot==0)continue; p[k++]=tot; tot=0; } } } int main() { while(gets(cmd)!=NULL) { read(); int i,j,t; for(i=0;i<k;i++) { if(i)printf(" "); printf("%d",p[i]); s[i]=p[i]; } printf("\n"); sort(s,s+k); for(i=k-1;i>0;i--) { if(p[i]==s[i])continue; for(j=1;j<i&&p[0]!=s[i];j++) { if(p[j]==s[i]) { printf("%d ",k-j); for(t=0;t<=j/2;t++) swap(p[t],p[j-t]); } } printf("%d ",k-i); for(t=0;t<=i/2;t++) swap(p[t],p[i-t]); } printf("0\n"); } return 0; } /* 8 4 6 7 5 2 1 5 6 2 3 4 */
3、110403/10037 Bridge (过桥)
贪心算法
1》最快和次快过桥,最快回来,最慢和次慢过桥,次快回来
使用时间2*p[0]+p[i]+p[i-1];
2》最快和最慢过桥,最快回来,最快和次慢过桥,最快回来
使用时间2*p[1]+p[0]+p[i];
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#include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> #include<ctype.h> using namespace std; int p[1005]; int rnd[2005][5]; int tot[2005]; void pass(int cnt,int t,int a,int b) { tot[cnt]=t; rnd[cnt][0]=p[a]; if(t>=2)rnd[cnt][1]=p[b]; } int main() { int T; scanf("%d",&T); while(T--) { int n,i; scanf("%d",&n); for(i=0;i<n;i++) { scanf("%d",&p[i]); } sort(p,p+n); int sum=0; int cnt=0; for(i=n-1;i>2;i-=2) { if(2*p[1]+p[0]+p[i]>2*p[0]+p[i]+p[i-1]) { sum+=2*p[0]+p[i]+p[i-1]; pass(cnt++,2,0,i-1); pass(cnt++,1,0,-1); pass(cnt++,2,0,i); pass(cnt++,1,0,-1); } else { sum+=2*p[1]+p[0]+p[i]; pass(cnt++,2,0,1); pass(cnt++,1,0,-1); pass(cnt++,2,i-1,i); pass(cnt++,1,1,-1); } } if(i==2) { sum+=p[0]+p[1]+p[2]; pass(cnt++,2,0,1); pass(cnt++,1,0,-1); pass(cnt++,2,0,2); } if(i==1) { sum+=p[1]; pass(cnt++,2,0,1); } if(i==0) { sum+=p[0]; pass(cnt++,1,0,-1); } printf("%d\n",sum); for(i=0;i<cnt;i++) { for(int j=0;j<tot[i];j++) { if(j)printf(" "); printf("%d",rnd[i][j]); } printf("\n"); } if(T)printf("\n"); } return 0; }
4、110404/10191 Longest Nap (最长打盹时间)
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#include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> #include<ctype.h> using namespace std; struct NODE { int s,e; }ap[105]; bool cmp(NODE a,NODE b) { return a.s<b.s; } int main() { int d; int day=1; while(scanf("%d",&d)!=EOF) { int i; char app[260]; int sh,eh,sm,em; for(i=0;i<d;i++) { scanf("%d:%d %d:%d",&sh,&sm,&eh,&em); gets(app); ap[i].s=sh*60+sm; ap[i].e=eh*60+em; } sort(ap,ap+d,cmp); int max=ap[0].s-10*60,tmp; int sta=10*60; for(i=1;i<d;i++) { tmp=ap[i].s-ap[i-1].e; if(max<tmp) { max=tmp; sta=ap[i-1].e; } } tmp=18*60-ap[d-1].e; if(max<tmp) { max=tmp; sta=ap[d-1].e; } printf("Day #%d: the longest nap starts at %d:%02d and will last for ",day++,sta/60,sta%60); if(max<60)printf("%d minutes.\n",max); else printf("%d hours and %d minutes.\n",max/60,max%60); } return 0; } /* 4 11:00 12:00 Lectures 12:00 13:00 Lunch, like always. 13:00 15:00 Boring lectures... 15:30 15:45 Reading */
5、110406/10138 CDVII (CDVII 高速公路)
注意点:1、没有消费的车不输出
2、采样的汽车位置可能exit的位置小于enter的位置
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#include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> #include<ctype.h> #include<map> #include<string> using namespace std; typedef long long lld; const int inf=1003; int fare[24]; int h[inf],nxt[inf]; struct PHOTO { char p[100]; }photo[inf]; map<string,int> m; struct NODE { int h,m; int dis; char w; }nod[inf]; bool cmp(PHOTO a,PHOTO b) { return strcmp(a.p,b.p)>0; } char name [inf][23]; typedef long long lld; int trans(char tmp[]) { int i,len=strlen(tmp),tot=0; for(i=0;i<len;i++) tot=tot*10+tmp[i]-'0'; return tot; } double calc(int t) { int i,dis,st; double tot=2; bool sta=true; for(i=h[t];i!=-1;i=nxt[i]) { if(nod[i].w==0) { dis=nod[i].dis; st=nod[i].h; sta=false; } if(nod[i].w==1&&!sta) { dis=abs(nod[i].dis-dis); tot+=dis*fare[st]*1.0/100+1; sta=true; } } return tot; } int main() { int T; scanf("%d",&T); while(T--) { int i,j; memset(h,-1,sizeof(h)); memset(nxt,-1,sizeof(nxt)); for(i=0;i<24;i++)scanf("%d",&fare[i]); int n=0; getchar(); while(gets(photo[n].p)!=NULL) { if(photo[n].p[0]=='\0')break; n++; } m.clear(); sort(photo,photo+n,cmp); map<string ,int>::iterator it; int tot=0; for(i=0;i<n;i++) { int len=strlen(photo[i].p),k=0; char tmp[100]=""; int cnt=0; int d; for(j=0;j<=len;j++) { if(j<len&&photo[i].p[j]!=' ') { tmp[k++]=photo[i].p[j]; } else { cnt++; tmp[k]='\0'; k=0; if(cnt==1) { it=m.find(tmp); if(it==m.end()) { m.insert(pair<string,int>(tmp,i)); strcpy(name[i],tmp); d=i; } else d=it->second; } else if(cnt==2) { nod[tot].h=(tmp[6]-'0')*10+tmp[7]-'0'; nod[tot].m=(tmp[9]-'0')*10+tmp[10]-'0'; } else if(cnt==3) { if(strcmp(tmp,"enter")==0) nod[tot].w=0; else nod[tot].w=1; } else { nod[tot].dis=trans(tmp); } } } nxt[tot]=h[d]; h[d]=tot; tot++; } for(it=m.begin();it!=m.end();it++) { int t=it->second; double x=calc(t); if(x>2) printf("%s $%.2f\n",name[it->second],x); } if(T)printf("\n"); } return 0; } /* 9 10 10 10 10 10 10 20 20 20 15 15 15 15 15 15 15 20 30 20 15 15 10 10 10 ABCD123 01:01:06:01 enter 17 765DEF 01:01:07:00 exit 95 ABCD123 01:01:08:03 exit 95 765DEF 01:01:05:59 enter 17 765DEF 01:01:04:01 enter 17 */
6、110407/10152 ShellSort (龟壳排序)
使用手工模拟很快能 出结果,即从 后往前遍历的最长子序列
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#include<stdio.h> #include<iostream> #include<string.h> #include<math.h> #include<algorithm> #include<ctype.h> #include<map> #include<string> using namespace std; char tur[205][1000]; char dest[205][1000]; map<string,int> m; map<int,string> mt; int arr[205]; int ind[205]; int main() { int T; scanf("%d",&T); while(T--) { int n,i,j; scanf("%d",&n); getchar(); m.clear(); mt.clear(); for(i=0;i<n;i++) { gets(tur[i]); } for(i=0;i<n;i++) { gets(dest[i]); m.insert(pair<string,int>(dest[i],i)); mt.insert(pair<int,string>(i,dest[i])); } map<string,int>::iterator it; map<int,string>::iterator itt; for(i=0;i<n;i++) { it=m.find(tur[i]); arr[i]=it->second; } int tot=0,cur=n-1; for(i=n-1;i>=0;i--) { if(arr[i]==cur) { tot++; cur--; } } for(i=n-1-tot;i>=0;i--) { itt=mt.find(i); cout<<itt->second<<endl; } printf("\n"); } return 0; }
也可以不用这么麻烦
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#include <stdio.h> #include<string.h> int main(void) { int n,i,ax,bx,num; scanf("%d",&n); //while(scanf("%d",&n)!=EOF) //{ while(n-->0) { char a[300][100],b[300][100]; scanf("%d",&num); getchar(); for(i=0;i<num;i++) gets(a[i]); for(i=0;i<num;i++) gets(b[i]); ax=num-1; bx=num-1; for(ax=num-1,bx=num-1;ax>=0;ax--) { if(!strcmp(a[ax],b[bx])) bx--; } while(bx>=0) printf("%s\n",b[bx--]); printf("\n"); } //} return 0; }
7、110408/10194 Football (aka Soccer) (足球)
注意:当平局的时候,以你不区分大小写的字典序排序
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#include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> #include<ctype.h> #include<string> #include<map> using namespace std; map<string,int> m; char tname[105],res[1005]; char team[35][100]; struct NODE { int no; int b,c,d,e,f,g,h,i; }node[35]; void calc(int ind,int sco,int ops) { node[ind].c++; node[ind].h+=sco; node[ind].i+=ops; if(sco==ops) { node[ind].e++; node[ind].b++; } else if(sco>ops) { node[ind].d++; node[ind].b+=3; } else node[ind].f++; } int getnum(char r[],int s,int e) { int i,tot=0; for(i=s;i<e;i++) tot=tot*10+r[i]-'0'; return tot; } void proc() { int i,len=strlen(res),s,e,mid,cnt=0,k=0,left,right,sl,sr; char tmp[100]; map<string,int>::iterator it; for(i=0;i<len;i++) { tmp[k]=res[i]; if(res[i]=='#'&&cnt==0) { tmp[k]='\0'; it=m.find(tmp); left=it->second; s=i; cnt++; } else if(res[i]=='#'&&cnt==1) { e=i; k=-1; } else if(res[i]=='@') { mid=i; } k++; } tmp[k]='\0'; it=m.find(tmp); right=it->second; sl=getnum(res,s+1,mid); sr=getnum(res,mid+1,e); calc(left,sl,sr); calc(right,sr,sl); } int _strcmp(char a[],char b[]) { int la=strlen(a),lb=strlen(b),i; for(i=0;i<la&&i<lb;i++) { char c=a[i]; char d=b[i]; if(c>='A'&&c<='Z')c=c-'A'+'a'; if(d>='A'&&d<='Z')d=d-'A'+'a'; if(c!=d)return c<d; } return la<lb; } bool cmp(NODE a,NODE b) { if(a.b!=b.b)return a.b>b.b; if(a.d!=b.d)return a.d>b.d; if(a.g!=b.g)return a.g>b.g; if(a.h!=b.h)return a.h>b.h; if(a.c!=b.c)return a.c<b.c; return _strcmp(team[a.no],team[b.no]); } int main() { int i,j,N,T,G; scanf("%d",&N); getchar(); while(N--) { gets(tname); scanf("%d",&T); getchar(); m.clear(); for(i=0;i<T;i++) { gets(team[i]); m.insert(pair<string,int>(team[i],i)); node[i].no=i; node[i].b=0; node[i].c=0; node[i].d=0; node[i].e=0; node[i].f=0; node[i].g=0; node[i].h=0; node[i].i=0; } scanf("%d",&G); getchar(); for(i=0;i<G;i++) { gets(res); proc(); } for(i=0;i<T;i++) { node[i].g=node[i].h-node[i].i; } sort(node,node+T,cmp); printf("%s\n",tname); for(i=0;i<T;i++) { printf("%d) %s %dp, %dg (%d-%d-%d), %dgd (%d-%d)\n", i+1,team[node[i].no],node[i].b,node[i].c,node[i].d,node[i].e,node[i].f,node[i].g,node[i].h,node[i].i); } if(N)printf("\n"); } return 0; } /* 3 World Cup 1998 - Group A 4 Brazil Norway Morocco Scotland 6 Brazil#14@1#Scotland Norway#2@14#Morocco Scotland#1@1#Norway Brazil#3@0#Morocco Morocco#3@0#Scotland Brazil#1@2#Norway Some strange tournament 5 Team A Team B aeam C Team D Team E 5 Team A#1@1#Team B Team A#2@2#aeam C Team A#0@0#Team D Team E#1@1#aeam C Team E#0@0#aeam C */
8 110405/10026 Shoemaker’s Problem (鞋匠的烦恼)
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#include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> #include<ctype.h> using namespace std; struct CUS { int no; double v; }cus[1005]; const double ex=1.0e-8; int dblcmp(double x) { if(fabs(x)<ex)return 0; return x>0?1:-1; } bool cmp(CUS a,CUS b) { return dblcmp(a.v-b.v)>0; } int main() { int T,N; scanf("%d",&T); while(T--) { scanf("%d",&N); int i,t,f; for(i=0;i<N;i++) { scanf("%d%d",&t,&f); cus[i].v=f*1.0/t; cus[i].no=i+1; } sort(cus,cus+N,cmp); for(i=0;i<N;i++) { if(i)printf(" "); printf("%d",cus[i].no); } printf("\n"); if(T)printf("\n"); } return 0; } /* 3 4 1 4 1 1000 1 2 1 5 4 2 10 1 100 1 1000 1 10000 4 1 10 2 100 1 1000 1 10000 4 1 100 1 200 2 10 1 10000 4 1 1 2 100 2 10 3 1 */