[題目鏈接]

https://www.lydsy.com/JudgeOnline/problem.php?id=2654

[算法]

給白色邊都加上一個值，會使得最小生成樹上的白邊數量減少，不妨二分給白色邊加上的值 , 檢驗答案時用Kruskal求最小生成樹即可

時間復雜度 : O(NlogN)

[代碼]

#include<bits/stdc++.h>
using namespace std;
#define MAXN 100010
#define MAXM 200010

struct edge
{
    int u , v , w , type;
} e[MAXN];

 
int tot , n , m , value , ans;
int fa[MAXN];

template <typename T> inline void chkmax(T &x , T y) { x = max(x , y); }
template <typename T> inline void chkmin(T &x , T y) { x = min(x , y); }
template <typename T> inline void read(T &x)
{
   T f = 1; x = 0;
   char c = getchar();
   
 
for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -f;
   for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - ‘0‘;
   x *= f;
}
inline int get_root(int x)
{
    if (fa[x] == x) return x;
    return fa[x] = get_root(fa[x]);
}
inline bool cmp(edge a , edge b)
{
    return a.w == b.w ? a.type < b.type : a.w < b.w;
}
inline 
 
bool check(int mid)
{
    int cnt = 0; 
    tot = 0;
    for (int i = 1; i <= n; i++) 
        fa[i] = i;
    for (int i = 1; i <= m; i++)
        if (!e[i].type) e[i].w += mid;    
    sort(e + 1 , e + m + 1 , cmp);
    for (int i = 1; i <= m; i++)
    {
        int su = get_root(e[i].u) , sv = get_root(e[i].v);
        if (su != sv)
        {
            fa[su] = sv;
            tot += e[i].w;
            if (!e[i].type) ++cnt;
        }
    }
    for (int i = 1; i <= m; i++) 
        if (!e[i].type) e[i].w -= mid;
    return cnt >= value;
}

int main()
{
    
    read(n); read(m); read(value);
    for (int i = 1; i <= m; i++)
    {
        read(e[i].u);
        read(e[i].v);
        read(e[i].w);
        ++e[i].u; ++e[i].v;
        read(e[i].type);    
    }
    int l = -105 , r = 105 , ans = 0;
    while (l <= r)
    {
        int mid = (l + r) / 2;
        if (check(mid))
        {
            ans = tot - value * mid;
            l = mid + 1;    
        } else r = mid - 1;
    }
    printf("%d\n" , ans);
    
    return 0;
}

[BZOJ 2654] tree