题目链接：
题目
D. Vitaly and Cycle
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
问题描述
After Vitaly was expelled from the university, he became interested in the graph theory.
Vitaly especially liked the cycles of an odd length in which each vertex occurs at most once.
Vitaly was wondering how to solve the following problem. You are given an undirected graph consisting of n vertices and m edges, not necessarily connected, without parallel edges and loops. You need to find t — the minimum number of edges that must be added to the given graph in order to form a simple cycle of an odd length, consisting of more than one vertex. Moreover, he must find w — the number of ways to add t edges in order to form a cycle of an odd length (consisting of more than one vertex). It is prohibited to add loops or parallel edges.
Two ways to add edges to the graph are considered equal if they have the same sets of added edges.
Since Vitaly does not study at the university, he asked you to help him with this task.
输入
The first line of the input contains two integers n and m ( — the number of vertices in the graph and the number of edges in the graph.
Next m lines contain the descriptions of the edges of the graph, one edge per line. Each edge is given by a pair of integers ai, bi (1 ≤ ai, bi ≤ n) — the vertices that are connected by the i-th edge. All numbers in the lines are separated by a single space.
It is guaranteed that the given graph doesn't contain any loops and parallel edges. The graph isn't necessarily connected.
输出
Print in the first line of the output two space-separated integers t and w — the minimum number of edges that should be added to the graph to form a simple cycle of an odd length consisting of more than one vertex where each vertex occurs at most once, and the number of ways to do this.
题意
给你一个无向图，问你最少添加几条边可以形成奇环，并且输出不同的方式数。
题解
最多只要添加三条边，所以我们可以分类讨论：
- 添加三条边
一条边都没有的时候
ans=C[n][3] - 添加两条边
每个顶点的度最大为1的时候
ans=m*(n-2) - 添加一条边
对每个连通块黑白染色，假设一个连通块黑的有b个，白的有w个，则：
ans+=b(b-1)/2+w(w-1)/2
代码
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
using namespace std;
const int maxn = 1e5 + 10;
const int maxm = maxn * 2;
typedef __int64 LL;
int n, m;
vector<int> G[maxn];
int deg[maxn];
int color[maxn];
void dfs(int u, int fa,LL &cnt,LL &r,LL &b) {
for (int i = 0; i < G[u].size(); i++) {
int v = G[u][i];
if (v == fa) continue;
if (!color[v]) {
color[v] = 3 - color[u];
if (color[v] == 1) r++;
else b++;
dfs(v, u, cnt, r, b);
}
else {
if (color[v] == color[u]) {
//printf("u:%d,v:%d\n", u, v);
cnt++;
}
}
}
}
int main() {
memset(deg, 0, sizeof(deg));
scanf("%d%d", &n, &m);
int maxv = -1;
for (int i = 0; i < m; i++) {
int u, v;
scanf("%d%d", &u,&v),u--,v--;
G[u].push_back(v);
G[v].push_back(u);
deg[u]++, deg[v]++;
}
for (int i = 0; i < n; i++) maxv = max(maxv, deg[i]);
if (m == 0) {
LL ans = (LL)n*(n - 1)*(n - 2) / 6;
printf("3 %I64d\n", ans);
return 0;
}
if (maxv < 2) {
LL ans = (LL)m*(n - 2);
printf("2 %I64d\n", ans);
return 0;
}
memset(color,0,sizeof(color));
LL cnt1 = 0, cnt2 = 0;
for (int i = 0; i < n; i++) {
LL r = 1, b = 0;
if (!color[i]) {
color[i] = 1;
dfs(i, -1,cnt1,r,b);
cnt2 += r*(r - 1) / 2 + b*(b - 1) / 2;
}
}
if (cnt1 == 0) {
printf("1 %I64d\n", cnt2);
}
else {
printf("0 %I64d\n", cnt1/2);
}
return 0;
}