BZOJ3206 [Apio2013]道路费用

首先我们强制要求几条待定价的边在MST中，建出MST

我们发现这个MST中原来的边是一定要被选上的，所以可以把点缩起来，搞成一棵只有$K$个点的树

然后$2^K$枚举每条边在不在最终的MST中，让在最终MST中的待定价的边尽量大，只需要在Kruskal的时候暴力更新每条边的定价即可

时间复杂度$O(m * logm + 2^K * K^2)$

 

  1 /**************************************************************
  2     Problem: 3206
  3     User: rausen
  4     Language: C++
  5     Result: Accepted
  6     Time:8040 ms
  7     Memory:8232 kb
  8 ****************************************************************/
  9  
 10 #include <cstdio>
 11 #include <algorithm>
 12  
 13 using namespace std;
 14 typedef long long ll;
 15 const int N = 1e5 + 5;
 16 const int M = 3e5 + 5;
 17 const int K = 25;
 18 const int inf = 1e9;
 19  
 20 inline int read();
 21  
 22 struct Edge {
 23     int x, y, v;
 24      
 25     inline void get(int f) {
 26         x = read(), y = read();
 27         if (f) v = read();
 28     }
 29      
 30     inline bool operator < (const Edge &E) const {
 31         return v < E.v;
 32     }
 33 } E[M], Ek[K], s[K];
 34  
 35 struct edge {
 36     int next, to;
 37     edge() {}
 38     edge(int _n, int _t) : next(_n), to(_t) {}
 39 } e[K << 1];
 40  
 41 int first[N], tot;
 42  
 43 struct tree_node {
 44     int fa, dep, mn;
 45     ll v, sum;
 46 } tr[N];
 47  
 48 int n, m, k, S, top;
 49 int fa[2][N];
 50 int root[K], cnt_root;
 51 int u[K];
 52 ll ans;
 53  
 54 inline void Add_Edges(int x, int y) {
 55     e[++tot] = edge(first[x], y), first[x] = tot;
 56     e[++tot] = edge(first[y], x), first[y] = tot;
 57 }
 58  
 59 int find(int x, int f) {
 60     return x == fa[f][x] ? x : fa[f][x] = find(fa[f][x], f);
 61 }
 62  
 63 #define y e[x].to
 64 void dp(int p) {
 65     int x;
 66     tr[p].sum = tr[p].v;
 67     for (x = first[p]; x; x = e[x].next)
 68         if (y != tr[p].fa) {
 69             tr[y].dep = tr[p].dep + 1, tr[y].fa = p;
 70             dp(y);
 71             tr[p].sum += tr[y].sum;
 72         }
 73 }
 74 #undef y
 75  
 76 ll work() {
 77     static int i, x, y, p;
 78     static ll res;
 79     for (tot = 0, i = 1; i <= k + 1; ++i) {
 80         p = root[i];
 81         fa[0][p] = p;
 82         first[p] = tr[p].fa = 0, tr[p].mn = inf;
 83     }
 84     for (i = 1; i <= k; ++i)
 85         if (u[i]) {
 86             x = find(Ek[i].x, 0), y = find(Ek[i].y, 0);
 87             if (x == y) return 0;
 88             fa[0][x] = y;
 89             Add_Edges(Ek[i].x, Ek[i].y);
 90         }
 91     for (i = 1; i <= k; ++i) {
 92         x = find(s[i].x, 0), y = find(s[i].y, 0);
 93         if (x != y) fa[0][x] = y, Add_Edges(s[i].x, s[i].y);
 94     }
 95     dp(S);
 96     for (i = 1; i <= k; ++i) {
 97         x = s[i].x, y = s[i].y;
 98         if (tr[x].dep < tr[y].dep) swap(x, y);
 99         while (tr[x].dep != tr[y].dep)
100             tr[x].mn = min(tr[x].mn, s[i].v), x = tr[x].fa;
101         while (x != y) {
102             tr[x].mn = min(tr[x].mn, s[i].v);
103             tr[y].mn = min(tr[y].mn, s[i].v);
104             x = tr[x].fa, y = tr[y].fa;
105         }
106     }
107 #define x Ek[i].x
108 #define y Ek[i].y
109     for (res = 0, i = 1; i <= k; ++i)
110         if (u[i])
111             res += tr[x].dep > tr[y].dep ? tr[x].mn * tr[x].sum : tr[y].mn * tr[y].sum;
112 #undef x
113 #undef y
114     return res;
115 }
116  
117 void dfs(int p) {
118     if (p == k + 1) {
119         ans = max(ans, work());
120         return;
121     }
122     u[p] = 0, dfs(p + 1);
123     u[p] = 1, dfs(p + 1);
124 }
125  
126 int main() {
127     int i, x, y;
128     n = read(), m = read(), k = read();
129     for (i = 1; i <= m; ++i) E[i].get(1);
130     for (i = 1; i <= k; ++i) Ek[i].get(0);
131     for (i = 1; i <= n; ++i) tr[i].v = read();
132     sort(E + 1, E + m + 1);
133     for (i = 1; i <= n; ++i) fa[0][i] = fa[1][i] = i;
134  
135     for (i = 1; i <= k; ++i)
136         fa[0][find(Ek[i].x, 0)] = find(Ek[i].y, 0);
137 #define x E[i].x
138 #define y E[i].y
139     for (i = 1; i <= m; ++i)
140         if (find(x, 0) != find(y, 0))
141             fa[0][find(x, 0)] = fa[0][find(y, 0)], fa[1][find(x, 1)] = fa[1][find(y, 1)];
142 #undef x
143 #undef y
144     S = find(1, 1);
145     for (i = 1; i <= n; ++i)
146         if (find(i, 1) != i) tr[find(i, 1)].v += tr[i].v;
147         else root[++cnt_root] = i;
148 #define x Ek[i].x
149 #define y Ek[i].y
150     for (i = 1; i <= k; ++i)
151         x = find(x, 1), y = find(y, 1);
152 #undef x
153 #undef y
154 #define x E[i].x
155 #define y E[i].y
156     for (i = 1; i <= m; ++i)
157         x = find(x, 1), y = find(y, 1);
158     for (i = 1; i <= m; ++i)
159         if (find(x, 1) != find(y, 1))
160             s[++top] = E[i], fa[1][find(x, 1)] = find(y, 1);
161 #undef x
162 #undef y
163     dfs(1);
164     printf("%lld\n", ans);
165     return 0;
166 }
167  
168 inline int read() {
169     static int x;
170     static char ch;
171     x = 0, ch = getchar();
172     while (ch < '0' || '9' < ch)
173         ch = getchar();
174     while ('0' <= ch && ch <= '9') {
175         x = x * 10 + ch - '0';
176         ch = getchar();
177     }
178     return x;
179 }
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posted on 2015-05-08 22:38  Xs酱~  阅读(965)  评论(0)  编辑  收藏  举报