CF 609E, 树链剖分

题目大意:给你一个联通无向图,问你包含某条边的最小生成树的大小是多少

解:做一个最小生成树,如果询问边在树上,则答案是最小生成树,否则则是这条边+树构成的环上去掉一条最大树边后得到的树。这里用树剖处理即可。

 

有个sb错误,因为问题是边权,而树剖的链一般是以点为单位,如果采用边权下放到点的技巧的话,注意lca的点权不要计算进来。

 

  1 #include <cstdio>
  2 #include <iostream>
  3 #include <algorithm>
  4 #include <utility>
  5 #include <map>
  6 #include <string>
  7 #include <cmath>
  8 #include <vector>
  9 #include <cstring>
 10 #include <stack>
 11 #include <set>
 12 #include <queue>
 13 
 14 using namespace std;
 15 
 16 #define SQR(x) ((x)*(x))
 17 #define LL long long
 18 #define LOWBIT(x) ((x)&(-(x)))
 19 #define PB push_back
 20 #define MP make_pair
 21 #define SQR(x) ((x)*(x))
 22 #define LSON(x) ((x)<<1)
 23 #define RSON(x) (((x)<<1)+1)
 24 
 25 
 26 
 27 #define MAXN 211111
 28 
 29 const double EPS = 1e-6;
 30 
 31 LL ans, spanTree;
 32 int n, m;
 33 
 34 struct line{
 35     int a, b, c, id;
 36     line(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), id(d) {}
 37     bool operator < (const line &rhs) const {
 38         return c < rhs.c;
 39     }
 40 };
 41 
 42 line a[MAXN];
 43 int w[MAXN];
 44 
 45 struct DisjointSet{
 46     int f[MAXN], n;
 47     void init(int nn = MAXN - 10) {
 48         n = nn;
 49         for (int i = 0; i <= n; ++i) {
 50             f[i] = i;
 51         }
 52     }
 53     int get(int x) {
 54         return f[x] == x ? x : f[x] = get(f[x]);
 55     }
 56     void joint(int x, int y) {
 57         f[x] = y;
 58     }
 59 } DjS;
 60 
 61 struct data{
 62     int dest, cost;
 63     data *nxt;
 64     data(int a = 0, int b = 0, data *c = 0): dest(a), cost(b), nxt(c) {}
 65 };
 66 
 67 struct TreeArray{
 68     int tree[MAXN], n;
 69     int a[MAXN];
 70     void init(int nn = MAXN - 1) {
 71         n = nn;
 72         memset(tree, 0, sizeof(tree[0])*(n+10));
 73     }
 74     void add(int x, int num = -1) {
 75         a[x] = num;
 76         while (x <= n) {
 77             tree[x] = max(tree[x], num);
 78             x += LOWBIT(x);
 79         }
 80     }
 81     int get(int x, int y) {
 82         int res = -1;
 83         while (x <= y) {
 84             if (y - LOWBIT(y) >= x) {
 85                 res = max(res, tree[y]);
 86                 y -= LOWBIT(y);
 87             } else {
 88                 res = max(res, a[y]);
 89                 --y;
 90             }
 91         }
 92         return res;
 93     }
 94 } TA;
 95 
 96 struct Tree{
 97     data edge[MAXN*2], *vect[MAXN];
 98     int n, cnt;
 99     int sz[MAXN], top[MAXN], heavy[MAXN], father[MAXN], dep[MAXN], lab[MAXN], num;
100     int root;
101     //???0?4??0?3??0?3??
102     void add(int x, int y, int z) {
103         edge[++cnt] = data(y, z, vect[x]); vect[x] = &edge[cnt];
104     }
105 
106     void init(int nn = MAXN - 10, int roott = 1) {
107         n = nn;
108         cnt = num = 0; 
109         memset(vect, 0, sizeof(vect[0])*(n+10));
110         memset(father, -1, sizeof(father[0])*(n+10));
111         root = roott;
112     }
113 
114     void split_dfs1(int u = 1) {
115         sz[u] = 1; heavy[u] = -1;
116         for (data *i = vect[u]; i; i = i->nxt) {
117             if (i->dest == father[u]) continue;
118             father[i->dest] = u;
119             dep[i->dest] = dep[u] + 1;
120             w[i->dest] = i->cost;
121             split_dfs1(i->dest);
122             sz[u] += sz[i->dest];
123             if (heavy[u] == -1 || sz[i->dest] > sz[heavy[u]]) heavy[u] = i->dest;
124         }
125     }
126 
127     void split_dfs2(int u = 1, int anc = 1) {
128         top[u] = anc; lab[u] = ++num;
129         TA.add(lab[u], w[u]);
130         if (heavy[u] != -1) split_dfs2(heavy[u], anc);
131         for (data *i = vect[u]; i; i = i->nxt) {
132             if (i->dest != father[u] && i->dest != heavy[u]) split_dfs2(i->dest, i->dest);
133         }
134     }
135 
136     void split() {
137         w[root] = -1;
138         dep[root] = 1;
139         father[root] = -1;
140         
141         split_dfs1(root);
142 
143         num = 0;
144 
145         split_dfs2(root, root);
146     }
147 
148     int query(int u, int v) {
149         int res = -1;
150         //cout << "query ================  " << ' '<< endl;
151         //cout << u << ' '<< v << endl;
152         //cout << top[u] <<' '<< top[v] << endl;
153         while (top[u] != top[v]) {
154             if (dep[top[u]] < dep[top[v]]) swap(u, v);
155             //todo
156         //    cout << u << ' '<< v << ' '<< top[u] << ' '<< top[v] << endl;
157             res = max(res, TA.get(lab[top[u]], lab[u]));
158         //    cout << res << endl;
159             //todo end
160             u = father[top[u]];
161         }
162         if (dep[u] > dep[v]) swap(u, v);
163         
164         //cout << u << ' '<< v << endl;
165         //cout << lab[u] << ' ' << lab[v] << endl;
166         //todo
167         res = max(res, TA.get(lab[u]+1, lab[v]));
168         //todo end
169         return res;
170     }
171 } G;
172 
173 void init() {
174     cin >> n >> m;
175     DjS.init(n);
176     G.init(n);
177     TA.init(n);
178 
179     int x, y, z;
180     for (int i = 0; i < m; ++i) {
181         cin >> x >> y >> z;
182         a[i] = line(x, y, z, i);
183     }
184     sort(a, a + m);
185     int cnt = 0;
186     spanTree = 0;
187     
188     for (int i = 0 ; i < m; ++i) {
189         int x = a[i].a, y = a[i].b;
190         x = DjS.get(x); y = DjS.get(y);
191         if (x != y) {
192             //cout << "=============" << endl;
193             //cout << a[i].a << ' ' << a[i].b << ' '<< a[i].c << endl;
194             DjS.joint(x, y);
195             spanTree += a[i].c;
196             ++cnt;
197             G.add(a[i].a, a[i].b, a[i].c);
198             G.add(a[i].b, a[i].a, a[i].c);
199             a[i].c *= -1;
200         }
201         if (cnt + 1 == n) break;
202     }
203     
204     G.split();
205     
206 }
207 
208 bool cmp2(const line &a, const line &b) {
209     return a.id < b.id;
210 }
211 
212 void solve() {
213     
214     //cout << spanTree << endl;
215     sort(a, a + m, cmp2);
216     ans = 0;
217     for (int i = 0; i < m; ++i) {
218         if (a[i].c < 0) {
219             ans = spanTree;
220         }
221         else {
222             ans = spanTree - G.query(a[i].a, a[i].b) + a[i].c;
223             //cout << a[i].a << ' ' << a[i].b << endl;
224             //cout << G.query(a[i].a, a[i].b) << ' '<< a[i].c << endl;
225         }
226         cout << ans << endl;
227     }
228 }
229 
230 int main() {
231     //freopen("test.txt", "r", stdin);
232     ios::sync_with_stdio(false);
233     init();
234     solve();
235     return 0;
236 }
CF 609E

 

posted @ 2016-09-24 09:21  F.D.His.D  阅读(...)  评论(...编辑  收藏