恩，归类上来讲的话。。。是一道非常好的noip题。。。只不过嘛、、、（此处省略100字）
然后将如何做：
首先Kruskal求出最小生成树。
我们其实可以发现严格的次小生成树不可能在MST上改两条边 <=> 只能改一条边。
那么如何改呢？
每次在MST中加入一条非树边，即不在MST的边，那么会形成一个环，只要找到换上的严格小于当前边权的最大值，删之，就形成了次小生成树的候选。
由Kruskal的算法保证加入的边权一定是环上最大的，因此我们要记录树链上的最大值和次大值（因为是严格小于）
而记录的方法就是倍增。。。noip难度。。。T T
对每条非树边都做一次即可。
复杂度大概是O(m * logm + n * logn + (m - n) * logn)
1 /**************************************************************
2 Problem: 1977
3 User: rausen
4 Language: C++
5 Result: Accepted
6 Time:1740 ms
7 Memory:32628 kb
8 ****************************************************************/
9
10 #include <cstdio>
11 #include <algorithm>
12
13 using namespace std;
14 typedef long long ll;
15 const int N = 100001;
16 const int M = 300001;
17 struct data{
18 int x, y, v;
19 bool selected;
20 }a[M];
21 struct edge{
22 int next, to ,v;
23 }e[N * 2];
24 struct tree_node{
25 int dep, fa[17], d1[17], d2[17];
26 }tr[N];
27 inline bool operator < (const data a, const data b){
28 return a.v < b.v;
29 }
30
31 int n, m, cnt, tot, del = 1e9;
32 int first[N], fa[N];
33 ll ans;
34
35 inline int read(){
36 int x = 0, sgn = 1;
37 char ch = getchar();
38 while (ch < ‘0‘ || ch > ‘9‘){
39 if (ch == ‘-‘) sgn = -1;
40 ch = getchar();
41 }
42 while (ch >= ‘0‘ && ch <= ‘9‘){
43 x = x * 10 + ch - ‘0‘;
44 ch = getchar();
45 }
46 return sgn * x;
47 }
48
49 inline void add_edge(int x, int y, int z){
50 e[++tot].next = first[x], first[x] = tot;
51 e[tot].to = y, e[tot].v = z;
52 }
53
54 void add_Edges(int X, int Y, int Z){
55 add_edge(X, Y, Z);
56 add_edge(Y, X, Z);
57 }
58
59 int find_fa(int x){
60 return x == fa[x] ? x : fa[x] = find_fa(fa[x]);
61 }
62
63 void dfs(int p){
64 int i, x, y, FA;
65 for (i = 1; i <= 16; ++i){
66 if (tr[p].dep < (1 << i)) break;
67 FA = tr[p].fa[i - 1];
68 tr[p].fa[i] = tr[FA].fa[i - 1];
69 tr[p].d1[i] = max(tr[p].d1[i - 1], tr[FA].d1[i - 1]);
70 if (tr[p].d1[i - 1] == tr[FA].d1[i - 1])
71 tr[p].d2[i] = max(tr[p].d2[i - 1], tr[FA].d2[i - 1]);
72 else {
73 tr[p].d2[i] = min(tr[p].d1[i - 1], tr[FA].d1[i - 1]);
74 tr[p].d2[i] = max(tr[p].d2[i - 1], tr[p].d2[i]);
75 tr[p].d2[i] = max(tr[p].d2[i], tr[FA].d2[i - 1]);
76 }
77 }
78 for (x = first[p]; x; x = e[x].next)
79 if ((y = e[x].to) != tr[p].fa[0]){
80 tr[y].fa[0] = p, tr[y].d1[0] = e[x].v, tr[y].dep = tr[p].dep + 1;
81 dfs(y);
82 }
83 }
84
85 inline int lca(int x, int y){
86 if (tr[x].dep < tr[y].dep) swap(x, y);
87 int tmp = tr[x].dep - tr[y].dep, i;
88 for (i = 0; i <= 16; ++i)
89 if ((1 << i) & tmp) x = tr[x].fa[i];
90 for (i = 16; i >= 0; --i)
91 if (tr[x].fa[i] != tr[y].fa[i])
92 x = tr[x].fa[i], y = tr[y].fa[i];
93 return x == y ? x : tr[x].fa[0];
94 }
95
96 void calc(int x, int f, int v){
97 int mx1 = 0, mx2 = 0, tmp = tr[x].dep - tr[f].dep, i;
98 for (i = 0; i <= 16; ++i)
99 if ((1 << i) & tmp){
100 if (tr[x].d1[i] > mx1)
101 mx2 = mx1, mx1 = tr[x].d1[i];
102 mx2 = max(mx2, tr[x].d2[i]);
103 x = tr[x].fa[i];
104 }
105 del = min(del, mx1 == v ? v - mx2 : v - mx1);
106 }
107
108 void work(int t, int v){
109 int x = a[t].x, y = a[t].y, f = lca(x, y);
110 calc(x, f, v);
111 calc(y, f, v);
112 }
113
114 int main(){
115 n = read(), m = read();
116 int i, f1, f2, TOT = 0;
117 for (i = 1; i <= m; ++i)
118 a[i].x = read(), a[i].y = read(), a[i].v = read();
119 for (i = 1; i <= n; ++i)
120 fa[i] = i;
121 sort(a + 1, a + m + 1);
122 for (i = 1; i <= m; ++i)
123 if ((f1 = find_fa(a[i].x)) != (f2 = find_fa(a[i].y))){
124 fa[f1] = f2;
125 ans += a[i].v;
126 a[i].selected = 1;
127 add_Edges(a[i].x, a[i].y, a[i].v);
128 ++TOT;
129 if (TOT == n - 1) break;
130 }
131 dfs(1);
132 for (i = 1; i <= m; ++i)
133 if (!a[i].selected)
134 work(i, a[i].v);
135 printf("%lld\n", ans + del);
136 return 0;
137 }
时间： 2024-12-26 17:07:46