洛谷P2619 Tree I

经典的k条白边MST

带权二分，按照套路我们要选择尽量少的白边。

 #include <cstdio>
 #include <algorithm>

 const int N = ;

 int D;

 struct Edge {
     int x, y, val, col;
     inline bool operator <(const Edge &w) const {
         if(val - D * col == w.val - D * w.col) {
             return col < w.col;
         }
         return val - D * col < w.val - D * w.col;
     }
 }edge[N];

 int n, m, k, ans;

 namespace ufs{
     int fa[N];
     int find(int x) {
         if(x == fa[x]) {
             return x;
         }
         return fa[x] = find(fa[x]);
     }
     inline void merge(int x, int y) {
         fa[find(x)] = find(y);
         return;
     }
     inline bool check(int x, int y) {
         return find(x) == find(y);
     }
     inline void clear() {
         for(int i = ; i <= n; i++) {
             fa[i] = i;
         }
         return;
     }
 }

 inline int check(int mid) {
     ufs::clear();
     D = mid;
     ans = ;
     int cnt = ;
     std::sort(edge + , edge + m + );
     for(int i = ; i <= m; i++) {
         if(!ufs::check(edge[i].x, edge[i].y)) {
             cnt += edge[i].col;
             ans += edge[i].val - D * edge[i].col;
             ufs::merge(edge[i].x, edge[i].y);
         }
     }
     return cnt;
 }

 int main() {
     scanf("%d%d%d", &n, &m, &k);
     for(int i = ; i <= m; i++) {
         scanf("%d%d%d%d", &edge[i].x, &edge[i].y, &edge[i].val, &edge[i].col);
         edge[i].x++;
         edge[i].y++;
         edge[i].col ^= ;
     }

     int l = -, r = ;
     while(l < r) {
         int mid = (l + r + ) >> ;
         int t = check(mid);
         //printf("[%d %d] mid = %d cnt = %d \n", l, r, mid, t);
         if(t == k) {
             printf("%d\n", ans + k * mid);
             return ;
         }
         if(t < k) {
             l = mid;
         }
         else {
             r = mid - ;
         }
     }
     check(r);
     printf("%d\n", ans + k * r);
     return ;
 }

AC代码