有一个trick就是没想到，枚举第二段时间后，要检测该火车能否继续跑一圈来判断，不能先检测前半圈能不能跑加进去后在检测后半段；
// **** 部分不能放在那个位置；
最近代码导致的错误总是找不出，贴下代码权当提醒吧！！
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<cmath> 5 #include<cstdlib> 6 #include<vector> 7 #include<algorithm> 8 using namespace std; 9 int us[180000]; 10 int ID[180000]; 11 vector<int> pr[5100]; 12 int ti[5100]; 13 int n; 14 void FIND(int i,int x,int nn,int y){ 15 int pt = ti[i] - x; 16 pr[nn].clear(); 17 pr[nn].push_back(ti[i] - x); 18 while ( true ) 19 { 20 int nt = pt + x; 21 if ( ID[nt] == -1 ) break; 22 us[nt] = 1; 23 // pr[nn].push_back(pt + x + y); **** 24 nt = nt + y + y; 25 if (ID[nt] == -1) break; 26 us[nt] = 1; 27 28 pr[nn].push_back(pt + x + y); 29 pr[nn].push_back(pt + x + y + y + x); 30 pt = pt + y * 2 + x + x; 31 } 32 } 33 int work(int x,int y) { 34 memset(us, 0, sizeof us); 35 pr[0].clear(); 36 int mm = 0; 37 us[ti[1]] = 1; 38 FIND(1,x,0,y); 39 int nn = 1; 40 mm = pr[0].size(); 41 if ( (mm - 1) % 2 ) return 0; 42 for ( int i = 2; i < n; i ++ ) 43 { 44 if ( us[ti[i]] ) continue; 45 us[i] = 1; 46 FIND(i,x,nn,y); 47 nn ++; 48 if ( pr[nn-1].size() != mm ) return false; 49 } 50 if (pr[0][2] <= pr[nn-1][0]) return 0; 51 52 for (int i = 0; i < nn; i++) { 53 int sz = pr[i].size(); 54 for (int j = 0; j < sz; j++) { 55 int hh,mm,ss; 56 hh = pr[i][j] / 3600; 57 mm = pr[i][j] % 3600 / 60; 58 ss = pr[i][j] % 3600 % 60; 59 printf("%02d:%02d:%02d%c",hh,mm,ss, j == sz-1 ? ' ':' '); 60 } 61 } 62 return 1; 63 } 64 void solve(){ 65 for (int k = 2; k < n; k++) { 66 int tx = ti[1] - ti[0], ty = ti[k] - ti[1]; 67 if (ty % 2) continue; 68 if (work(tx,ty/2)) return; 69 } 70 // cout<<" ** "<< endl; 71 } 72 int main(){ 73 // freopen("in.txt","r",stdin); 74 n = 0; 75 int hh,mm,ss; 76 memset(ID,-1,sizeof(ID)); 77 while (~scanf("%d:%d:%d",&hh,&mm,&ss)) { 78 ti[n++] = hh * 3600 + mm * 60 + ss; 79 ID[ti[n-1]] = n-1; 80 } 81 // cout<<" ** "<<endl; 82 solve(); 83 return 0; 84 }