Codeforces Round #311 (Div. 2) D. Vitaly and Cycle (two-figure dyeing, singular ring)

Given n points, M-bar edges.

The minimum number of edges required to make the odd ring in the diagram, and the output of the method of adding edges at this time

According to test instructions, it is only possible to

0: The odd ring already exists, DFS has been dyed and the color is the same

1: Determine the number of stained black and white in each connected block

2: The degree of a point > 1

3:0 Sides

1#include <cstdio>2#include <iostream>3#include <queue>4#include <vector>5#include <stack>6#include <Set>7#include <string>8#include <cstring>9#include <math.h>Ten#include <algorithm> One#include <map> A  -#include <sstream> -#include <ctype.h> the  - #defineLson l,m,rt<<1 - #defineRson m+1,r,rt<<1|1 -   +typedefLong Longll; - using namespacestd; +   A Const intINF =0x7f7f7f7f; at Const DoublePI = ACOs (-1.0); - Const intMAXN =1005; - Const intMOD = (int) 1e9+7; - #defineLOCAL 0 - #defineMOD 1000000007 -  in  - /* to 0 Sides: 0 1 + 1 Sides: - 2-side m* (n-2) the 3 side Cn3 * */ $ intn,m;Panax Notoginseng intdegree[100005]; -vector<int> g[100005]; the intcolor[100005]; + intw[100005],b[100005];  A BOOLf,f0; the intCnt//several unicom blocks + voidDfsintUintCintnum) { -Color[u] =C; $     if(C &1) w[num]++; $     Elseb[num]++; -      -      for(inti =0; I < g[u].size (); i + +){ the         intv =G[u][i]; -         if(!Color[v]) {WuyiDFS (V,3-c,num); the}Else{ -             if(Color[u] = =Color[v]) { WuF0 =true; -             } About         } $     } - } -  - voidsolve () { A     //3 sides +     if(M = =0){ theprintf"%d%i64d\n",3, 1ll*n* (n1) * (n2)/6); -         return ; $     } the      for(inti =0; I <= N; i + +) g[i].clear (); thememset (Degree,0,sizeof(degree)); thef =true; the      for(inti =0; I < m; i + +){ -         intu,v; inscanf"%d%d",&u,&v); theG[u].push_back (v); degree[u]++; theG[v].push_back (U); degree[v]++; About         if((Degree[u) >1) || (Degree[v] >1)) F =false; the     } the     //2 sides the     if(f) { +printf"2%i64d\n", 1ll*m* (n2)); -         return ; the     }  Bayi      theF0 =false; theCNT =0; -memset (Color,0,sizeof(color)); -Memset (W,0,sizeof(w)); thememset (b,0,sizeof(b)); the      for(inti =1; I <= N; i + +){ the         if(!Color[i]) theDFS (I,1, cnt++); -     } the      the     //0 Sides the     if(F0) {94printf"0 1\n"); the         return; the     } the     98     //1 Sides Aboutll ans =0;  -      for(inti =0; I < CNT; i + +){101Ans = ans + 1ll*w[i]* (w[i]-1)/2+ 1ll*b[i]* (b[i]-1)/2;102     }103printf"1%i64d\n", ans);104     return ; the }106 intMain () {107     108      while(SCANF ("%d%d", &n,&m)! =EOF) {109 solve (); the     }111      the     return 0;113}

Codeforces Round #311 (Div. 2) D. Vitaly and Cycle (two-figure dyeing, singular ring)