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kruskal
kruskal跑出最小生成树
对图中每条边进行讨论(设这条边连接的两个点为
u
,
v
u,v
u,v)
- 非树边,要使这条边成为树边,把这条边的权值最大改为 u − > v u->v u−>v路径上边权最大值-1即可
- 树边,最大边权使得这条边依然是树边,最大能达到所有连接的点的路径经过这条边的非树边的边权最小值-1
如何做?非树边和树边分别做一下
需要维护路径最值,路径修改最值
转化成裸树剖了。。。
20
W
20W
20W的范围两只
l
o
g
log
log问题不大
码量较大,调半天。。。
Code:
#include <bits/stdc++.h>
#define maxn 200010
#define inf 0x3f3f3f3f
#define ls rt << 1
#define rs rt << 1 | 1
using namespace std;
struct Edge{
int to, next, len, id;
}edge[maxn << 1];
struct Line{
int x, y, z, flag, id;
}line[maxn << 1];
struct Seg{
int l, r, Max, Min, tag;
}seg[maxn << 2];
int n, m, num, head[maxn], fa[maxn], f[maxn], d[maxn], size[maxn], son[maxn], id[maxn], Index;
int pre[maxn], prelen[maxn], wt[maxn], top[maxn], ans[maxn];
inline int read(){
int s = 0, w = 1;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
return s * w;
}
void addedge(int x, int y, int z, int id){ edge[++num] = (Edge){ y, head[x], z, id }; head[x] = num; }
bool cmp(Line x, Line y){ return x.z < y.z; }
int get(int k){ return k == f[k] ? k : f[k] = get(f[k]); }
void kruskal(){
for (int i = 1; i <= n; ++i) f[i] = i;
for (int i = 1; i <= m; ++i){
int s1 = get(line[i].x), s2 = get(line[i].y);
if (s1 != s2){
f[s1] = s2, line[i].flag = 1;
addedge(line[i].x, line[i].y, line[i].z, line[i].id);
addedge(line[i].y, line[i].x, line[i].z, line[i].id);
}
}
}
void dfs(int u){
size[u] = 1, son[u] = -1;
for (int i = head[u]; i; i = edge[i].next){
int v = edge[i].to;
if (v != fa[u]){
d[v] = d[u] + 1, fa[v] = u, pre[v] = edge[i].id, prelen[v] = edge[i].len;
dfs(v);
size[u] += size[v];
if (son[u] == -1 || son[u] != -1 && size[son[u]] < size[v]) son[u] = v;
}
}
}
void dfs(int u, int x){
wt[id[u] = ++Index] = prelen[u], top[u] = x;
if (son[u] == -1) return;
dfs(son[u], x);
for (int i = head[u]; i; i = edge[i].next){
int v = edge[i].to;
if (v != fa[u] && v != son[u]) dfs(v, v);
}
}
void updmax(int &x, int y){ if (x < y) x = y; }
void updmin(int &x, int y){ if (x > y) x = y; }
void pushup(int rt){
seg[rt].Max = max(seg[ls].Max, seg[rs].Max);
seg[rt].Min = min(seg[ls].Min, seg[rs].Min);
}
void pushdown(int rt){
updmin(seg[ls].Min, seg[rt].tag);
updmin(seg[ls].tag, seg[rt].tag);
updmin(seg[rs].Min, seg[rt].tag);
updmin(seg[rs].tag, seg[rt].tag);
seg[rt].tag = inf;
}
void build(int rt, int l, int r){
seg[rt].Min = seg[rt].tag = inf;
seg[rt].l = l, seg[rt].r = r;
if (l == r){ seg[rt].Max = wt[l]; return; }
int mid = (l + r) >> 1;
build(ls, l, mid); build(rs, mid + 1, r);
pushup(rt);
}
void update(int rt, int l, int r, int k){
if (seg[rt].l > r || seg[rt].r < l) return;
if (seg[rt].l >= l && seg[rt].r <= r){
updmin(seg[rt].Min, k);
updmin(seg[rt].tag, k);
return;
}
pushdown(rt);
update(ls, l, r, k); update(rs, l, r, k);
pushup(rt);
}
int querymax(int rt, int l, int r){
if (seg[rt].l > r || seg[rt].r < l) return 0;
if (seg[rt].l >= l && seg[rt].r <= r) return seg[rt].Max;
pushdown(rt);
return max(querymax(ls, l, r), querymax(rs, l, r));
}
int querymin(int rt, int pos){
if (seg[rt].l == seg[rt].r) return seg[rt].Min;
pushdown(rt);
if (seg[ls].r >= pos) return querymin(ls, pos); else
return querymin(rs, pos);
}
int qrymax(int u, int v){
int sum = 0;
while (top[u] != top[v]){
if (d[top[u]] < d[top[v]]) swap(u, v);
updmax(sum, querymax(1, id[top[u]], id[u]));
u = fa[top[u]];
}
if (d[u] > d[v]) swap(u, v);
return max(sum, querymax(1, id[u] + 1, id[v]));
}
void updatemin(int u, int v, int k){
while (top[u] != top[v]){
if (d[top[u]] < d[top[v]]) swap(u, v);
update(1, id[top[u]], id[u], k);
u = fa[top[u]];
}
if (d[u] > d[v]) swap(u, v);
update(1, id[u] + 1, id[v], k);
}
int main(){
n = read(), m = read();
for (int i = 1; i <= m; ++i) line[i].x = read(), line[i].y = read(), line[i].z = read(), line[i].id = i;
sort(line + 1, line + 1 + m, cmp);
kruskal();
dfs(1); dfs(1, 1);
build(1, 1, n);
for (int i = 1; i <= m; ++i)
if (!line[i].flag){
ans[line[i].id] = qrymax(line[i].x, line[i].y) - 1;
updatemin(line[i].x, line[i].y, line[i].z);
}
for (int i = 2; i <= n; ++i) ans[pre[i]] = querymin(1, id[i]) - 1;
for (int i = 1; i <= m; ++i) printf("%d ", ans[i] < inf - 1 ? ans[i] : -1);
return 0;
}

本文介绍了一种利用Kruskal算法求解最小生成树,并通过树剖技术优化路径查询与更新的方法。详细解释了如何处理图中的树边与非树边,以达到路径最值的快速维护。
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