这个题加深了我对主席树的理解,是个好题。每次更新某个点的距离时,是以之前对这个点的插入操作形成的线段树为基础,在O(logn)的时间中造出了一颗新的线段树,相比直接创建n颗线段树更省时间。比较的时候二分比较,为了加快比较给每个点设置一个hash值。
代码:
#include <bits/stdc++.h>
using namespace std;
const unsigned long long P = 13331;
const int mod = 1000000007;
const int maxn = 100010;
int head[maxn], Next[maxn * 2], edge[maxn * 2], ver[maxn * 2];
int tot, tote;
int b[maxn * 2], pre[maxn];
int mx, dis[maxn];
bool vis[maxn];
int Stack[maxn], Top;
void add(int x, int y, int z) {
ver[++tote] = y;
edge[tote] = z;
Next[tote] = head[x];
head[x] = tote;
}
struct SegementTree {
int ls, rs;
int sum;
unsigned long long hash;
}tr[maxn * 200];
void pushup(int x) {
tr[x].sum = tr[tr[x].ls].sum + tr[tr[x].rs].sum;
unsigned long long tmp1 = tr[tr[x].ls].hash, tmp2 = tr[tr[x].rs].hash;
tr[x].hash = (((tmp1 * P + tmp2) ^ tmp1) * P ^ tmp2) * P;
}
void insert(int &now, int l, int r, int pos) {
int p = now;
now = ++tot;
tr[now] = tr[p];
if(l == r) {
tr[now].sum = 1;
tr[now].hash = l + P;
return;
}
int mid = (l + r) >> 1;
if(pos <= mid) insert(tr[now].ls, l, mid, pos);
else insert(tr[now].rs, mid + 1, r, pos);
pushup(now);
}
void del(int &now, int l, int r, int ql, int qr) {
if(l >= ql && r <= qr) {
now = 0;
return;
}
int p = now;now = ++tot;
tr[now] = tr[p];
int mid = (l + r) >> 1;
if(ql <= mid) del(tr[now].ls, l, mid, ql, qr);
if(qr > mid) del(tr[now].rs, mid + 1, r, ql, qr);
pushup(now);
}
int query(int now, int l, int r, int pos) {
if(l == r) {
if(tr[now].sum == 0) return l;
return -1;
}
int mid = (l + r) >> 1;
if(pos > mid) return query(tr[now].rs, mid + 1, r, pos);
int ans = query(tr[now].ls, l, mid, pos);
if(ans != -1) return ans;
return query(tr[now].rs, mid + 1, r, pos);
}
int add(int now, int pos) {
int p = query(now, 0, mx, pos);
if(p > pos) del(now, 0, mx, pos, p - 1);
insert(now, 0, mx, p);
return now;
}
int get_sum(int now, int l, int r) {
int mid = (l + r) >> 1;
if(now == 0) return 0;
if(l == r) return b[l];
return (get_sum(tr[now].ls, l, mid) + get_sum(tr[now].rs, mid + 1, r)) % mod;
}
bool cmp(int x, int y, int l, int r) {
if(l == r) return tr[x].sum >= tr[y].sum;
int mid = (l + r) >> 1;
if(tr[tr[x].rs].hash != tr[tr[y].rs].hash)
return cmp(tr[x].rs, tr[y].rs, mid + 1, r);
else return cmp(tr[x].ls, tr[y].ls, l, mid);
}
struct node {
int x, y;
bool operator < (const node& rhs) const {
return cmp(x, rhs.x, 0, mx);
}
};
priority_queue<node> q;
void dijkstra(int s) {
dis[s] = ++tot, q.push((node){dis[s], s});
while(!q.empty()) {
node tmp = q.top();
q.pop();
if(vis[tmp.y]) continue;
vis[tmp.y] = 1;
int x = tmp.y;
for (int i = head[x]; i; i = Next[i]) {
int y = ver[i], z = add(dis[x], edge[i]);
if(!dis[y] || !cmp(z, dis[y], 0, mx)) {
pre[y] = x;
dis[y] = z;
q.push((node){z, y});
}
}
}
}
void print(int x, int deep) {
if(x == 0) {
printf("%d\n", deep);
return;
}
print(pre[x], deep + 1);
printf("%d ", x);
}
int main() {
int n, m, x, y, z, s, t;
scanf("%d%d", &n, &m);
for (int i = 1; i <= m; i++) {
scanf("%d%d%d", &x, &y, &z);
add(x, y, z);
add(y, x, z);
mx = max(mx, z);
}
b[0] = 1;
mx += 20;
for (int i = 1; i <= mx; i++) {
b[i] = (b[i - 1] << 1) % mod;
}
scanf("%d%d", &s, &t);
dijkstra(s);
if(dis[t] == 0) printf("-1\n");
else {
printf("%d\n", get_sum(dis[t], 0, mx));
while(t) {
Stack[++Top] = t;
t = pre[t];
}
printf("%d\n", Top);
while(Top) {
printf("%d ", Stack[Top]);
Top--;
}
}
}
Codeforces 464E The Classic Problem (最短路 + 主席树 + hash)
原文:https://www.cnblogs.com/pkgunboat/p/10566988.html