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 }
View Code

 

By Xs酱~ 转载请说明 博客地址:http://www.cnblogs.com/rausen
posted on 2014-10-31 22:30  Xs酱~  阅读(209)  评论(0编辑  收藏  举报