bzoj 1977

题意:求严格的次小生成树。点n<=100000,m<=300000

思路:很容易想到先做一边最小生成树,然后枚举每条非树边(u, v, w),然后其实就是把u,v路径上小于w的最大边替换成w,对于所有的这种新树取一个权值最小的即可。。

       然后就变成求u,v的最大值及次大值。。树链剖分和lct显然是可以做的。。

       不过很早就知道倍增却一直没写过,今天就正好写一发。。

 

code:

  1 /**************************************************************
  2     Problem: 1977
  3     User: yzcstca
  4     Language: C++
  5     Result: Accepted
  6     Time:2344 ms
  7     Memory:35048 kb
  8 ****************************************************************/
  9 
 10 #include<cstdio>
 11 #include<iostream>
 12 #include<cstring>
 13 #include<cstdlib>
 14 #include<cmath>
 15 #include<algorithm>
 16 #include<string>
 17 #include<map>
 18 #include<set>
 19 #include<vector>
 20 #include<queue>
 21 #include<stack>
 22 #include<ctime>
 23 #define repf(i, a, b) for (int i = (a); i <= (b); ++i)
 24 #define M0(x)  memset(x, 0, sizeof(x))
 25 #define vii vector< pair<int, int> >::iterator
 26 #define x first
 27 #define y second
 28 #define two(i) (1 << i)
 29 using namespace std;
 30 typedef long long ll;
 31 typedef pair<int, int> pii;
 32 const int maxn = 101001;
 33 const int maxm = 301000;
 34 struct oo{
 35     int u, v, w;
 36     bool operator<(const oo& p) const{
 37         return w < p.w;
 38     }
 39 } E[maxm];
 40 vector<pii> e[maxn];
 41 int fa[maxn], intree[maxm];
 42 int n, m, ans;
 43 int dep[maxn], f[maxn][20], vv[maxn][20][2];
 44  
 45 void init(){
 46     for (int i = 0; i < m; ++i)
 47          scanf("%d%d%d", &E[i].u, &E[i].v, &E[i].w);
 48     repf(i, 0, n) e[i].clear();
 49 }
 50  
 51  
 52 inline int find(const int& k){
 53     return fa[k] == k ? k : fa[k] = find(fa[k]);
 54 }
 55  
 56 int vis[maxn], t[maxn];
 57 void bfs(){
 58      M0(vis);
 59      queue<int> q;
 60      q.push(1), dep[1] = 0, vis[1] = 1;
 61      int u, v, cnt;
 62      t[cnt = 1] = 1;
 63      while (!q.empty()){
 64           u = q.front();
 65           q.pop();
 66           for (vii it = e[u].begin(); it != e[u].end(); ++it){
 67                   if (vis[it->x]) continue;
 68                   dep[it->x] = dep[u] + 1;
 69                   f[it->x][0] = u;
 70                   vv[it->x][0][0] = it->y, vv[it->x][0][1] = -1;
 71                   q.push(it->x), vis[it->x] = 1, t[++cnt] = it->x;
 72           }
 73      }
 74 }
 75  
 76 void update(int v[],const int& val){
 77      if (val > v[0]) 
 78           swap(v[0], v[1]), v[0] = val;
 79      else if (val < v[0] && val > v[1])
 80           v[1] = val;
 81 }
 82  
 83 void rmq(){
 84     int u, v;
 85     repf(i, 1, n){
 86          u = t[i];
 87          for (int i = 1; i <= 17; ++i){
 88               if (two(i) > dep[u]) break;
 89               v = f[u][i-1];
 90               f[u][i] = f[v][i-1];
 91               vv[u][i][0] = vv[u][i][1] = -1;
 92               update(vv[u][i], vv[u][i-1][0]);
 93               update(vv[u][i], vv[u][i-1][1]);
 94               update(vv[u][i], vv[v][i-1][0]);
 95               update(vv[u][i], vv[v][i-1][1]);
 96          }
 97     }
 98 }
 99  
100 int res[2];
101 void query(int u, int s, int res[]){
102     for (int i = 0; i <= 17; ++i) if (s & two(i))
103          update(res, vv[u][i][0]), update(res, vv[u][i][1]), u = f[u][i], s ^= two(i);
104 }
105  
106 void work(int u, int v, const int& w){
107      if (dep[u] > dep[v]) swap(u, v);
108      int fu = u, fv = v, h;
109      if (dep[fu] != dep[fv]){
110             h = dep[fv] - dep[fu];
111             for (int i = 0; i <= 17; ++i) if (h & two(i))
112                   h ^= two(i), fv = f[fv][i];
113      }
114      if (fu == fv){
115              res[0] = res[1] = -1;
116              query(v, dep[v] - dep[u], res);
117              h = (res[0] != w) ? res[0] : res[1];
118              if (h != -1) ans = min(ans, w - h);
119              return;
120      }
121      for (int i = 17; i >= 0; --i){
122             if (two(i) > dep[fu]) continue;
123             if (f[fu][i] != f[fv][i])
124                   fu = f[fu][i], fv = f[fv][i];
125      }
126      fu = f[fu][0];
127      res[0] = res[1] = -1;
128      query(u, dep[u] - dep[fu], res);
129      query(v, dep[v] - dep[fu], res);
130      h = (res[0] != w) ? res[0] : res[1];
131      if (h != -1) ans = min(ans, w - h);
132 }
133  
134 void solve(){
135     repf(i, 0, n) fa[i] = i;
136     sort(E, E + m);
137     memset(intree, 0, sizeof(int) * (m + 10));
138     int u, v, fu, fv, w;
139     ll mst = 0;
140     repf(i, 0, m-1){
141          u = E[i].u, v = E[i].v, w = E[i].w;
142          fu = find(u), fv = find(v);
143          if (fu != fv){
144              fa[fu] = fv, intree[i] = 1;
145              mst += E[i].w;
146              e[u].push_back(make_pair(v, w));
147              e[v].push_back(make_pair(u, w) );
148          }
149     }
150     bfs();
151     rmq();
152     ans = 0x3fffffff;
153     for (int i = 0; i < m; ++i) if (!intree[i])
154          work(E[i].u, E[i].v, E[i].w);
155     cout << mst + ans << endl;
156          
157 }
158  
159 int main(){
160 //    freopen("a.in", "r", stdin);
161 //    freopen("a.out", "w", stdout);
162     while (scanf("%d%d", &n, &m) != EOF){
163            init();
164            solve();
165     }
166     return 0;
167 }
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posted on 2014-11-03 22:30  yzcstc  阅读(286)  评论(0编辑  收藏  举报