AcWing 257. Explanation of prisoner detention (bipartite picture)

AcWing 257. Detain criminals
Their thinking ： Use two points to find the smallest mid value , The two points of the edge larger than this value need to be divided into two parts of the bipartite graph , That is to dye into two different colors . All points are dyed , Judge whether it is caused by conflict , If any, increase mid, If there is no conflict, we can continue to reduce mid, Because a bipartite graph with reduced edges must still be a bipartite graph

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 20010, M = 200010;

int n, m;
int h[N], e[M], w[M], ne[M], idx;
int color[N];

void add(int a, int b, int c)
{
    
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++ ;
}

bool dfs(int u, int c, int mid)
{
    
    color[u] = c;
    for (int i = h[u]; ~i; i = ne[i])
    {
    
        int j = e[i];
        if (w[i] <= mid) continue;
        if (color[j])
        {
    
            if (color[j] == c) return false;
        }
        else if (!dfs(j, 3 - c, mid)) return false;
    }

    return true;
}

bool check(int mid)
{
    
    memset(color, 0, sizeof color);
    for (int i = 1; i <= n; i ++ )
        if (!color[i])
            if (!dfs(i, 1, mid))
                return false;
    return true;
}

int main()
{
    
    scanf("%d%d", &n, &m);
    memset(h, -1, sizeof h);

    while (m -- )
    {
    
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        add(a, b, c), add(b, a, c);
    }

    int l = 0, r = 1e9;
    while (l < r)
    {
    
        int mid = l + r >> 1;
        if (check(mid)) r = mid;  // What is required in the question mid The minimum value of , When mid establish ,mid The values on the right are also true , but mid-1 Not necessarily , therefore r=mid 
        else l = mid + 1;  // If this value does not hold , Then we need to find a larger value  
    }

    printf("%d\n", r);

    return 0;
}