UOJ #117. 欧拉回路

#117. 欧拉回路

在图中找一个环使得每条边都在环上出现恰好一次。


要注意的地方好多啊

每条边恰好出现一次!!!


条件:每个点偶度 / 入度=出度

方法就是套圈法啦


然后本题自环是合法的,如果20000个(1,1)边的话会被卡成\(O(n^2)\),所以加上当前弧优化

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
#define fir first
#define sec second
const int N = 2e5+5, inf = 1e9+5;
inline int read() {
    char c=getchar(); int x=0,f=1;
    while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
    return x*f;
}

int type, n, m, u, v;
struct edge {int v, ne;} e[N<<1];
int cnt = 1, h[N];

namespace undir {
    inline void ins(int u, int v) {
        e[++cnt] = (edge) {v, h[u]}; h[u] = cnt;
        e[++cnt] = (edge) {u, h[v]}; h[v] = cnt;
    }
    int de[N], st[N], top, mark[N<<1]; 
    void dfs(int u) { //if(++t % 1000 == 0) printf("t %d\n", t);
        for(int &i=h[u]; i; i=e[i].ne) if(!mark[i]) {
            mark[i] = mark[i^1] = 1;
            int j = i;
            dfs(e[i].v);
            st[++top] = j&1 ? -((j-1)>>1) : j>>1; 
        }
    }
    void solve() {
        for(int i=1; i<=m; i++) u = read(), v = read(), ins(u, v), de[u]++, de[v]++;
        for(int i=1; i<=n; i++) if(de[i] & 1) {puts("NO"); return;}
        int u = 1;
        while(u <=n && !de[u]) u++; 
        dfs(u);
        if(top != m) {puts("NO"); return;}
        puts("YES");
        while(top) printf("%d ", st[top--]);
    }
}
namespace dir {
    inline void ins(int u, int v) {
        e[++cnt] = (edge) {v, h[u]}; h[u] = cnt;
    }
    int ind[N], outd[N], st[N], top, mark[N];
    void dfs(int u) {
        for(int &i=h[u]; i; i=e[i].ne) if(!mark[i]) {
            mark[i] = 1;
            int j = i;
            dfs(e[i].v);
            st[++top] = j-1;
        }
    }
    void solve() {
        for(int i=1; i<=m; i++) u = read(), v = read(), ins(u, v), outd[u]++, ind[v]++;
        for(int i=1; i<=n; i++) if(ind[i] != outd[i]) {puts("NO"); return;}
        int u = 1;
        while(u <=n && !outd[u]) u++; 
        dfs(u);
        if(top != m) {puts("NO"); return;}
        puts("YES");
        while(top) printf("%d ", st[top--]);
    }
}
int main() {
    freopen("in.in", "r", stdin);
    type = read();
    n = read(); m = read();
    if(type == 1) undir::solve();
    else dir::solve();
}
posted @ 2017-05-15 15:48 Candy? 阅读(...) 评论(...) 编辑 收藏