标签：std detail oid lin int func cin rac lang
给出\(N\)个点的简单无向图，不一定联通，现在要用最少的路径去覆盖所有边，并且每条边只被覆盖一次，问最少路径覆盖数和各条路径
\(N\le 10^5\)
对于每个连通块分别处理
考虑每个联通块，必然是用最少的欧拉路径去覆盖，首先考虑连通块里没有奇数度数的点的情况，这个情况下只要跑欧拉回路即可
如果连通块中有\(x\)个奇数度数的点，那么显然\(2|x\)，且必然是用\(\frac{x}{2}\)条欧拉路径去覆盖，每两个奇数度数的顶点之间会有一条欧拉路径，考虑如何构造路径，首先将奇数度数的顶点两两配对连边，只剩下一对奇数度数点不连边，然后在新建的图中跑欧拉路径（此时必然存在欧拉路径），可以发现其中\(\frac{x}{2}-1\)条新加入的边正好把路径分成了\(\frac{x}{2}\)条，这些分开来的路径正好是所求路径
//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 1e5+7;
int n, m, deg[MAXN], bel[MAXN], vis[MAXN<<2], num[MAXN];
vector<int> pt[MAXN];
struct Graph{
int head[MAXN],to[MAXN<<2],nxt[MAXN<<2],tot,id[MAXN<<2];
void clear(){ tot = 0; memset(head,255,MAXN<<2); }
void addEdge(int u, int v, int idd){
to[tot] = v; nxt[tot] = head[u]; id[tot] = idd;
vis[tot] = false; head[u] = tot++;
to[tot] = u; nxt[tot] = head[v]; id[tot] = -idd;
vis[tot] = false; head[v] = tot++;
}
}G;
void mark(int u, int id){
bel[u] = id;
pt[id].push_back(u);
for(int i = G.head[u]; ~i; i = G.nxt[i]){
int v = G.to[i];
if(!bel[v]) mark(v,id);
}
}
stack<int> stk;
void euler(int u){
int now = ++num[u];
for(int i = G.head[u]; ~i; i = G.nxt[i]){
if(vis[i]) continue;
G.head[u] = G.nxt[i];
vis[i] = vis[i^1] = true;
euler(G.to[i]);
stk.push(G.id[i]);
if(now!=num[u]) break;
}
}
void print(){
printf("%d ",stk.size());
while(!stk.empty()){
printf("%d%c",stk.top()," \n"[stk.size()==1]);
stk.pop();
}
}
void rua(int id){
int odddeg = 0;
for(int &x : pt[id]) if(deg[x]&1) odddeg++;
if(!odddeg){
euler(pt[id][0]);
print();
}
else{
int last = -1;
for(int &x : pt[id]){
if(odddeg==2) break;
if(deg[x]&1){
if(last==-1) last = x;
else{
G.addEdge(last,x,0);
deg[last]++; deg[x]++;
last = -1;
odddeg -= 2;
}
}
}
for(int &x : pt[id]) if(deg[x]&1) last = x;
euler(last);
vector<int> vec;
while(true){
vec.clear();
while(!stk.empty() and stk.top()!=0){
vec.push_back(stk.top());
stk.pop();
}
if(!stk.empty()) stk.pop();
printf("%d",vec.size());
for(int x : vec) printf(" %d",x);
puts("");
if(stk.empty()) break;
}
}
}
void solve(){
G.clear();
memset(deg+1,0,n<<2);
for(int i = 1; i <= m; i++){
int u, v; scanf("%d %d",&u,&v);
G.addEdge(u,v,i);
deg[u]++, deg[v]++;
}
int ID = 0;
memset(bel+1,0,n<<2);
int __count = 0;
for(int i = 1; i <= n; i++) if(!bel[i]){
pt[++ID].clear();
mark(i,ID);
if(pt[ID].size() > 1){
int odddeg = 0;
for(int x : pt[ID]) if(deg[x]&1) odddeg++;
if(!odddeg) __count++;
else __count += odddeg / 2;
}
}
printf("%d\n",__count);
for(int i = 1; i <= ID; i++){
if(pt[i].size() == 1) continue;
rua(i);
}
}
int main(){
while(scanf("%d %d",&n,&m)!=EOF) solve();
return 0;
}
标签：std detail oid lin int func cin rac lang
原文地址：https://www.cnblogs.com/kikokiko/p/13189907.html
