SGU121 Bridges painting

题目大意

给出一张无向图，对每条边01染色
要求对于每个度数大于1的顶点，连出去的边中两种颜色都有
构造染色方案

算法思路

不难发现，不存在的情况一定是独立的奇环
故首先从度数不为2的点出发，进行交错染色
注意在同一顶点多次出发时，初始的颜色应当不同

简单的证明：
路径上的点由于前后两边的颜色不同，故只需考虑度数大于2的起点u，不妨设以0出发
1. 没有回到u，则下次以1出发
2. 以1回到u，符合要求
3. 以0回到u，由于度数大于2，以1继续出发

对于剩下的环，任选起点进行染色即可

时间复杂度: O(V+E)

代码

/** * Copyright © 2015 Authors. All rights reserved. *  * FileName: 121.cpp * Author: Beiyu Li <[e-mail]> * Date: 2015-05-31 */#include <bits/stdc++.h>using namespace std;#define rep(i,n) for (int i = 0; i < (n); ++i)#define For(i,s,t) for (int i = (s); i <= (t); ++i)#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)typedef long long LL;typedef pair<int, int> Pii;const int inf = 0x3f3f3f3f;const LL infLL = 0x3f3f3f3f3f3f3f3fLL;const int maxn = 100 + 5;int n, psz;struct Edge {        int v;        Edge *next;} epl[maxn*maxn], *e[maxn];int deg[maxn];int color[maxn][maxn];void add_edge(int u, int v){        Edge *i = epl + psz++;        i->v = v; i->next = e[u]; e[u] = i;        ++deg[u];}void dfs(int u, int c){        for (Edge *i = e[u]; i; i = i->next) {                int v = i->v;                if (color[u][v]) continue;                color[u][v] = color[v][u] = c;                dfs(v, c = 3 - c);        }}void print(int u, Edge *i){        if (!i) return;        print(u, i->next); printf("%d ", color[u][i->v]);}bool solve(){        rep(u,n) if (deg[u] != 2) dfs(u, 1);        rep(u,n) dfs(u, 1);        rep(u,n) {                int c = 0;                for (Edge *i = e[u]; i; i = i->next) c |= color[u][i->v];                if (c != 3 && deg[u] > 1) return false;        }        rep(u,n) print(u, e[u]), puts("0");        return true;}int main(){        scanf("%d", &n);        rep(u,n) {                int v;                while (scanf("%d", &v) && v) add_edge(u, --v);        }        if (!solve()) puts("No solution");        return 0;}

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