這題主要是個套路。。就是求無向圖最小路徑覆蓋。。

與有向圖的二分圖做法不同，這個是轉化為求最少的尤拉路徑。。

尤拉圖有個結論是尤拉路徑的個數為度為奇數的點的個數/2（可以類比歐拉回路的結論）

然後求尤拉路徑的方法是fleury演算法。。其思想就是暴力dfs，然後巧妙的地方就是邊是方向取的，即以出棧的順序為尤拉路徑。。

然後就是一大堆細節問題。。大概是今天沒什麼人做出來的原因。。。

這題其實覆蓋得情況比較全面（偶數的歐拉回路，奇數的歐拉回路和尤拉路徑、分塊求尤拉路徑等），作為模板其實挺合適。。可以記下來。。

/**
 *　　　　　　  　　┏┓　　 　┏┓
 * 　　　　　  　　┏┛┗━━━━━━━┛┗━━━┓
 * 　　　　　  　　┃　　　　　　　┃ 　
 * 　　　　　  　　┃　　　━　　 　┃
 * 　　　　　  　　┃　＞　　　＜　┃
 * 　　　　　  　　┃　　　　　　　┃
 * 　　　　　  　　┃...　⌒　... 　┃
 * 　　　　　　  　┃              ┃
 * 　　　　　　  　┗━┓          ┏━┛
 * 　　　　　　　　　┃          ┃　Code is far away from bug with the animal protecting　　　　　　　　　　
 * 　　　　　　　　　┃          ┃   神獸保佑,程式碼無bug
 * 　　　　　　　　　┃          ┃　　　　　　　　　　　
 * 　　　　　　　　　┃          ┃  　　　　　　
 * 　　　　　　　　　┃          ┃
 * 　　　　　　　　　┃          ┃　　　　　　　　　　　
 * 　　　　　　　　　┃          ┗━━━┓
 * 　　　　　　　　　┃              ┣┓
 * 　　　　　　　　　┃              ┏┛
 * 　　　　　　　　　┗┓┓┏━━━━━━━━┳┓┏┛
 * 　　　　　　　　　　┃┫┫       ┃┫┫
 * 　　　　　　　　　　┗┻┛       ┗┻┛
 */ 
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-12
#define succ(x) (1LL<<x)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define mid (x+y>>1)
#define NM 100005
#define nm 400000
#define N 1000005
#define M(x,y) x=max(x,y)
const double pi=acos(-1);
const ll inf=998244353;
using namespace std;
ll read(){
    ll x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return f*x;
}



struct edge{int t,v;bool f;edge*next,*rev;}e[nm],*h[NM],*o=e;
void _add(int x,int y,int v){o->t=y;o->v=v;o->next=h[x];h[x]=o++;}
void add(int x,int y,int v){_add(x,y,v);_add(y,x,-v);h[x]->rev=h[y];h[y]->rev=h[x];}
int n,m,_x,_y,cnt,b[NM],s;
bool v[NM];
vector<int>a[NM];

void dfs(int x){
    v[x]=true;
    link(x)if(!j->f){
	j->rev->f=j->f=true;
	dfs(j->t);
	if(j->v==0)a[++cnt].clear();else a[cnt].push_back(-j->v);
    }
}

int main(){
    while(~scanf("%d%d",&n,&m)){
	mem(e);mem(h);o=e;mem(b);s=0;mem(v);cnt=0;
	inc(i,1,m){
	    _x=read();_y=read();
	    add(_x,_y,i);
	    b[_x]++;b[_y]++;
	}
	_x=0;
	inc(i,1,n)if(b[i]&1){
		if(_x)add(_x,i,0),_x=0;
		else _x=i;
	}
	inc(i,1,n)if(!v[i]&&b[i]%2)a[++cnt].clear(),dfs(i),cnt--;
	inc(i,1,n)if(!v[i]&&b[i])a[++cnt].clear(),dfs(i);
	printf("%d\n",cnt);
	inc(i,1,cnt){printf("%d ",m=a[i].size());inc(j,0,m-2)printf("%d ",a[i][j]);printf("%d\n",a[i][m-1]);}
	//printf(":::\n");
    }
    return 0;
}
 

Cover

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 225    Accepted Submission(s): 36
Special Judge
 

Problem Description

The Wall has down and the King in the north has to send his soldiers to sentinel.
The North can be regard as a undirected graph (not necessary to be connected), one soldier can cover one path. Today there's no so many people still breathing in the north, so the King wants to minimize the number of soldiers he sent to cover each edge exactly once. As a master of his, you should tell him how to arrange soldiers.

Input

There might be multiple test cases, no more than 20. You need to read till the end of input.
In the first line, two integers n and m, representing the number of nodes and edges in the graph.
In the following m lines, each contain two integers, representing two ends of an edge.
There are no parallel edges or self loops.
1≤n

Output

For each test case, the first line contains number of needed routes, p.
For the following p lines, an integer x in the beginning, followed by x integers, representing the list of used edges. Every integer should be a positive or negative integer. Its absolute value represents the number of chosen edge (1~n). If it's positive, it shows that this edge should be passed as the direction as the input, otherwise this edge should be passed in the direction different from the input. Edges should be in correct order.

Sample Input

3 3 1 2 1 3 2 3

Sample Output

1 3 1 3 -2

Source

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