8.图的邻点不同颜色

class Solution {
public:
    vector<int> gardenNoAdj(int N, vector<vector<int>>& paths) {
        vector<int> G[N];
        for (int i=0; i<paths.size(); i++){//建立邻接表
            G[paths[i][0]-1].push_back(paths[i][1]-1);
            G[paths[i][1]-1].push_back(paths[i][0]-1);
        }
        vector<int> answer(N,0);//初始化全部未染色
        for(int i=0; i<N; i++){//对每个结点
            set<int> color{1,2,3,4};
            for (int j=0; j<G[i].size(); j++){//当前结点的邻接点
                color.erase(answer[G[i][j]]);//把已染过色的去除
            }
            answer[i]=*(color.begin());//染色
        }
        return answer;
    }
};

　　Johnkram 版

class Solution {
public:
    int h[10005],ne[40005],p[40005],n,m,d[10005],f[10005];
    bool v[10005];
    void dfs(int x)
    {
        v[x]=1;
        for(int i=h[x];i;i=ne[i])if(!v[p[i]])
        {
            f[p[i]]=x;
            d[p[i]]=d[x]+1;//深度层
            dfs(p[i]);
        }
    }
    vector<int> gardenNoAdj(int N, vector<vector<int>>& paths) {
        n=paths.size();
        int i,j;
        for(i=m=0;i<n;i++)
        {
            p[++m]=paths[i][1];
            ne[m]=h[paths[i][0]];
            h[paths[i][0]]=m;
            p[++m]=paths[i][0];
            ne[m]=h[paths[i][1]];
            h[paths[i][1]]=m;
        }
        n=N;
        vector<int> ans;
        for(i=1;i<=n;i++)if(!v[i])dfs(i);//防多图
        for(i=1;i<=n;i++)ans.push_back(d[i]%2+1);
        for(i=1;i<=n;i++)for(j=h[i];j;j=ne[j])if(d[p[j]]<d[i]&&p[j]!=f[i]&&ans[i-1]==ans[p[j]-1])ans[p[j]-1]=d[p[j]]%2+3;//防有环
        return ans;
    }
};

给跪了。两个%2+1,和%2+3，思路实在是太清晰了！！