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。

 1 #include <cstdio>
 2 #include <algorithm>
 3 
 4 const int N = 100010;
 5 
 6 int D;
 7 
 8 struct Edge {
 9     int x, y, val, col;
10     inline bool operator <(const Edge &w) const {
11         if(val - D * col == w.val - D * w.col) {
12             return col < w.col;
13         }
14         return val - D * col < w.val - D * w.col;
15     }
16 }edge[N];
17 
18 int n, m, k, ans;
19 
20 namespace ufs{
21     int fa[N];
22     int find(int x) {
23         if(x == fa[x]) {
24             return x;
25         }
26         return fa[x] = find(fa[x]);
27     }
28     inline void merge(int x, int y) {
29         fa[find(x)] = find(y);
30         return;
31     }
32     inline bool check(int x, int y) {
33         return find(x) == find(y);
34     }
35     inline void clear() {
36         for(int i = 1; i <= n; i++) {
37             fa[i] = i;
38         }
39         return;
40     }
41 }
42 
43 inline int check(int mid) {
44     ufs::clear();
45     D = mid;
46     ans = 0;
47     int cnt = 0;
48     std::sort(edge + 1, edge + m + 1);
49     for(int i = 1; i <= m; i++) {
50         if(!ufs::check(edge[i].x, edge[i].y)) {
51             cnt += edge[i].col;
52             ans += edge[i].val - D * edge[i].col;
53             ufs::merge(edge[i].x, edge[i].y);
54         }
55     }
56     return cnt;
57 }
58 
59 int main() {
60     scanf("%d%d%d", &n, &m, &k);
61     for(int i = 1; i <= m; i++) {
62         scanf("%d%d%d%d", &edge[i].x, &edge[i].y, &edge[i].val, &edge[i].col);
63         edge[i].x++;
64         edge[i].y++;
65         edge[i].col ^= 1;
66     }
67 
68     int l = -150, r = 150;
69     while(l < r) {
70         int mid = (l + r + 1) >> 1;
71         int t = check(mid);
72         //printf("[%d %d] mid = %d cnt = %d \n", l, r, mid, t);
73         if(t == k) {
74             printf("%d\n", ans + k * mid);
75             return 0;
76         }
77         if(t < k) {
78             l = mid;
79         }
80         else {
81             r = mid - 1;
82         }
83     }
84     check(r);
85     printf("%d\n", ans + k * r);
86     return 0;
87 }