【F. Graph Without Long Directed Paths】贪心_f. graph without long directed paths 题意-优快云博客

该博客探讨了如何构建有向图，确保图中不存在路径长度超过2。通过贪心策略，从一个节点开始染色，并确保相邻节点染相反颜色。如果遇到已染相同颜色的节点，则无法构建满足条件的图，输出否定结果。

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F
F. Graph Without Long Directed Paths

题意给你n个点 m条边保证联通问你能否构造一张有向图(不一定联通)使得路径长度不超过2

我们很容易想到从一个点开始染色和他相连的点染相反颜色然后再继续递归下去染色

但是要判断一下这个点染色和下个点染色是否相反如果那个点已经有颜色和自己是一样的那么肯定是NO

例如这组数据

实际上是不能染色的因为一定会有一条2的路径所以你把相同颜色出现的时候全局flag为false 输出NO即可

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;
 
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
 
typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;
 
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
const int MAX_N = 2e5+5;
int col[MAX_N],a[MAX_N],b[MAX_N];
vector<int> G[MAX_N];
bool flag;
void dfs(int x,int fa,int v)
{
    int sz = G[x].size();
    col[x] = v;
    for(int i = 0;i<sz;++i)
    {
        int to = G[x][i];
        if(to==fa) continue;
        if(col[to]==-1) dfs(to,x,1-v);
        else if(col[to] == col[x]) flag = false;
    }
    return ;
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,m;
    flag = true;
    scanf("%d%d",&n,&m);
    memset(col,-1,sizeof(col));
    for(int i = 1;i<=m;++i)
    {
        scanf("%d%d",&a[i],&b[i]);
        G[a[i]].push_back(b[i]);
        G[b[i]].push_back(a[i]);
    }
    dfs(1,-1,0);
    if(!flag)
    {
        printf("NO\n");
        return 0;
    }
    printf("YES\n");
    for(int i = 1;i<=m;++i)
    {
        if(col[a[i]]==1) printf("1");
        else printf("0");
    }
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}