「模拟测试」管道-倍增

给出一个无向图,对于每条边,求必须选择这条边的情况下,使得原图连通的最小代价。

题解

先建出最小生成树,要选择 $(u, v)$ 这条边,若已在树中直接输出,不在树中,倍增找到路径 $(u, v)$ 上权值最大的边,然后用这条边替换掉就是答案。

时间复杂度 $O(m \log n)$。

代码

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/**
* Copyright (c) 2017, xehoth
* All rights reserved.
* 「SuperOJ 2020」管道 30-10-2017
* 倍增
* @author xehoth
*/
#include <bits/stdc++.h>
namespace IO {
inline char read() {
static const int IN_LEN = 1000000;
static char buf[IN_LEN], *s, *t;
s == t ? t = (s = buf) + fread(buf, 1, IN_LEN, stdin) : 0;
return s == t ? -1 : *s++;
}
template <typename T>
inline bool read(T &x) {
static char c;
static bool iosig;
for (c = read(), iosig = false; !isdigit(c); c = read()) {
if (c == -1) return false;
c == '-' ? iosig = true : 0;
}
for (x = 0; isdigit(c); c = read()) x = x * 10 + (c ^ '0');
iosig ? x = -x : 0;
return true;
}
inline void read(char &c) {
while (c = read(), isspace(c) && c != -1)
;
}
inline int read(char *buf) {
register int s = 0;
register char c;
while (c = read(), isspace(c) && c != -1)
;
if (c == -1) {
*buf = 0;
return -1;
}
do
buf[s++] = c;
while (c = read(), !isspace(c) && c != -1);
buf[s] = 0;
return s;
}
const int OUT_LEN = 1000000;
char obuf[OUT_LEN], *oh = obuf;
inline void print(char c) {
oh == obuf + OUT_LEN ? (fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf) : 0;
*oh++ = c;
}
template <typename T>
inline void print(T x) {
static int buf[30], cnt;
if (x == 0) {
print('0');
} else {
x < 0 ? (print('-'), x = -x) : 0;
for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 | 48;
while (cnt) print((char)buf[cnt--]);
}
}
inline void print(const char *s) {
for (; *s; s++) print(*s);
}
inline void flush() { fwrite(obuf, 1, oh - obuf, stdout); }
struct InputOutputStream {
template <typename T>
inline InputOutputStream &operator>>(T &x) {
read(x);
return *this;
}
template <typename T>
inline InputOutputStream &operator<<(const T &x) {
print(x);
return *this;
}
~InputOutputStream() { flush(); }
} io;
}
const int MAXN = 100000;
const int MAXM = MAXN * 2;
namespace UnionFindSet {
int fa[MAXN + 1];
inline int get(const register int x) {
return x == fa[x] ? x : fa[x] = get(fa[x]);
}
}
namespace {
using IO::io;
using UnionFindSet::get;
const int MAX_LOG = 17;
bool vis[MAXM + 1];
struct Edge {
int u, v, w, id;
inline bool operator<(const Edge &p) const { return w < p.w; }
} edges[MAXM + 1];
struct Node {
int v, w;
Node(int v, int w) : v(v), w(w) {}
};
std::vector<Node> edge[MAXN + 1];
typedef std::vector<Node>::iterator Iterator;
inline void addEdge(const int u, const int v, const int w) {
edge[u].push_back(Node(v, w)), edge[v].push_back(Node(u, w));
}
int fa[MAX_LOG][MAXN + 1], max[MAX_LOG][MAXN + 1], dep[MAXN + 1];
long long ans[MAXM + 1];
inline int optMax(const register int a, const register int b) {
return a > b ? a : b;
}
void dfs(const int u) {
dep[u] = dep[fa[0][u]] + 1;
for (register int i = 1; (1 << i) <= dep[u]; i++) {
fa[i][u] = fa[i - 1][fa[i - 1][u]];
max[i][u] = optMax(max[i - 1][u], max[i - 1][fa[i - 1][u]]);
}
for (register Iterator p = edge[u].begin(); p != edge[u].end(); p++)
if (p->v != fa[0][u]) fa[0][p->v] = u, max[0][p->v] = p->w, dfs(p->v);
}
inline int getMax(register int u, register int v) {
register int ret = INT_MIN;
dep[u] < dep[v] ? std::swap(u, v) : (void)0;
register int delta = dep[u] - dep[v];
for (register int i = 0; (1 << i) <= delta; i++)
if (delta & (1 << i)) ret = optMax(ret, max[i][u]), u = fa[i][u];
if (u == v) return ret;
for (register int i = MAX_LOG - 1; i >= 0; i--) {
if (fa[i][u] != fa[i][v]) {
ret = optMax(ret, max[i][u]), u = fa[i][u];
ret = optMax(ret, max[i][v]), v = fa[i][v];
}
}
return optMax(ret, optMax(max[0][u], max[0][v]));
}
inline void solve() {
register int n, m;
io >> n >> m;
for (register int i = 1; i <= n; i++) UnionFindSet::fa[i] = i;
for (register int i = 1; i <= m; i++)
io >> edges[i].u >> edges[i].v >> edges[i].w, edges[i].id = i;
std::sort(edges + 1, edges + m + 1);
register long long cost = 0;
for (register int i = 1, cnt = 0, u, v; i <= m; i++) {
if ((u = get(edges[i].u)) != (v = get(edges[i].v))) {
UnionFindSet::fa[u] = v, cost += edges[i].w;
addEdge(edges[i].u, edges[i].v, edges[i].w);
vis[i] = true;
if (++cnt == n - 1) break;
}
}
dfs(1);
for (register int i = 1; i <= m; i++) {
if (vis[i]) {
ans[edges[i].id] = cost;
continue;
}
ans[edges[i].id] = cost - getMax(edges[i].u, edges[i].v) + edges[i].w;
}
for (register int i = 1; i <= m; i++) io << ans[i] << '\n';
}
}
int main() {
// freopen("sample/1.in", "r", stdin);
solve();
return 0;
}
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