Codeforces 508 D. Tanya and Password (Fleury algorithm)

Codeforces 508 D. Tanya and Password (Fleury algorithm)

Topic Links:

Http://codeforces.ru/problemset/problem/508/D

Test instructions
Give n a string of length 3, such as: ABC bca AAB if the suffix of a string is 2 and the length of the other string is a prefix of 2, then the two strings can be connected, for example: AABCA, and then the N strings can form a graph, and a Oraton path on the graph is obtained.
Limit:
1 <= N <= 2*10^5, with uppercase letters in the string, lowercase letters
Ideas:
Each string is treated as an edge, and its prefix suffix is used as a point, such as: ABC, AB to BC.
Then the problem is: there is a 62*62 point, the 2*10^5 edge of the map, the question of a European pull path.
With Fleury (Frolles), the personal feeling uses adjacency table to achieve efficiency ratio adjacency matrix

PS: This problem DFS will explode stack, change it to non-recursive on the line.

Attach Fleury algorithm link

Http://www.cnblogs.com/Lyush/archive/2013/04/22/3036659.html

C + + Code

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/*codeforces 508 D. Tanya and Password Test instructions Give n a string of length 3, such as: ABC bca AAB if the suffix of a string is 2 and the length of the other string is a prefix of 2, then the two strings can be connected, for example: AABCA, and then the N strings can form a graph, and a Oraton path on the graph is obtained. Limit: 1 <= N <= 2*10^5, with uppercase letters in the string, lowercase letters Ideas: Each string is treated as an edge, and its prefix suffix is used as a point, such as: ABC, AB to BC. Then the problem is: there is a 62*62 point, the 2*10^5 edge of the map, the question of a European pull path. With Fleury (Frolles), the personal feeling uses adjacency table to achieve efficiency ratio adjacency matrix PS: This problem DFS will explode stack, change it to non-recursive on the line. */ #include<iostream> #include<cstdio> #include<vector> #include<string> #include<cstring> #include<map> #include<stack> usingnamespaceStd Constintm=200005; Constintn= $* $; vector<int> Mp[n]; intFa[n]; intIn[n],out[n]; CharANS[M],REV[M]; intlen=0; String I2s[n]; Map<string,int> s2i; intstk[m],top=0; intGETFA (intx) { if(X!=fa[x])returnFA[X]=GETFA (Fa[x]); returnX } voidDfsintx) { while(Mp[x].size () >0){ Stk[++top]=x; intCh=mp[x][mp[x].size ()-1]; Mp[x].pop_back (); X=ch; } Stk[++top]=x; } voidFleury (intS) { intBrige; Stk[++top]=s; intflag=0; intTp while(top>0){ tp=stk[top--]; Brige= (mp[tp].size () = =0); if(Brige) { if(flag) ans[len++]=i2s[tp][0]; Else{ ans[len++]=i2s[tp][1]; ans[len++]=i2s[tp][0]; flag=1; } } Else DFS (TP); } } intMain () { intN,m; Charstr[5]; String fr,to; n=0; scanf"%d", &m); for(intI=0;i< $* $; ++i) Fa[i]=i; for(intI=0; i<m;++i) { intx, y; intFx,fy; scanf'%s ', str); fr=str[0]; fr+=str[1]; to=str[1]; to+=str[2]; X=S2I[FR]; if(x==0{s2i[fr]=++n; i2s[n]=fr; x=n;} Y=s2i[to]; if(y==0{s2i[to]=++n; i2s[n]=to; y=n;} Mp[x].push_back (y); FX=GETFA (x); FY=GETFA (y); if(FX!=FY) FA[FY]=FX; ++OUT[X]; ++in[y]; } intlt=1; intROOT=GETFA (1); for(intI=2; i<=n;++i) { if(GETFA (i)!=root) {lt=0; Break; } } ints=0, t=0, cnt=0; for(intI=1; i<=n;++i) { if(In[i]==out[i])Continue; Else{ cnt++; if(in[i]-out[i]==1) t=i; Elseif(in[i]-out[i]==-1) S=i; Else{cnt=-1; Break; } } } if(LT && (cnt==2&& s!=t) | | cnt==0)){ Puts"YES"); if(cnt==0) s=1; Fleury (S); intLen=strlen (ANS); for(intI=0; i<len;++i) { rev[len-1-i]=ans[i]; } Puts (rev.); } Else{ Puts"NO"); } return0; }

Codeforces 508 D. Tanya and Password (Fleury algorithm)