我测得的是这个东西在不开\(O2\)的情况下比\(pair + priority_queue\)快了将近\(1/3\)
#include<bits/stdc++.h>
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
using namespace std;
const int N1 = 1e5 + 10, N2 = 2e5 + 10;
const int INF = 0x3f3f3f3f;
int head[N1], to[N2], nex[N2], value[N2], cnt;
int dis[N1];
int n, m, s;
inline int read(){
int x = 0,f=1;char ch = getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
while(ch<=‘9‘&&ch>=‘0‘){x=x*10+ch-‘0‘;ch=getchar();}
return x*f;
}
struct point {
int id, dis;
point(int a = 0, int b = 0) : id(a), dis(b) {}
}a[N1 << 2];
point minp(point a, point b) {
if(a.dis < b.dis) return a;
return b;
}
void build_tree(int rt, int l, int r) {
if(l == r) {
a[rt].id = l;
a[rt].dis = INF;
return ;
}
build_tree(lson);
build_tree(rson);
a[rt].dis = INF;
a[rt].id = a[rt << 1].id;
}
void update_tree(int rt, int l, int r, int dis, int x) {
if(l == r) {
a[rt].dis = x;
return ;
}
if(dis <= mid) update_tree(lson, dis, x);
else update_tree(rson, dis, x);
a[rt] = minp(a[rt << 1], a[rt << 1 | 1]);
}
void add(int x, int y, int w) {
to[cnt] = y;
nex[cnt] = head[x];
value[cnt] = w;
head[x] = cnt++;
}
void print_tree(int rt, int l, int r) {
printf("%d %d\n", rt, a[rt].id);
if(l == r) {
// printf("%d %d\n", a[rt].id, a[rt].dis);
return ;
}
print_tree(lson);
print_tree(rson);
}
void Dijkstra() {
memset(dis, 0x3f, sizeof dis);
dis[s] = 0;
// print_tree(1, 1, n);
update_tree(1, 1, n, 1, 0);
// puts("");
// print_tree(1, 1, n);
for(int cas = 1; cas < n; cas++) {
// cout << 1 << endl;
point temp = a[1];
// printf("%d %d\n", temp.id, temp.dis);
update_tree(1, 1, n, temp.id, INF);
for(int i = head[temp.id]; ~i; i = nex[i]) {
if(dis[to[i]] > dis[temp.id] + value[i]) {
// cout << 1 << endl;
dis[to[i]] = dis[temp.id] + value[i];
update_tree(1, 1, n, to[i], dis[to[i]]);
}
}
}
for(int i = 1; i <= n; i++)
printf("%d%c", dis[i], i == n ? ‘\n‘ : ‘ ‘);
}
int main() {
// freopen("in.txt", "r", stdin);
int x, y, w;
memset(head, -1, sizeof head);
n = read(), m = read(), s = read();
build_tree(1, 1, n);
for(int i = 0; i < m; i++) {
x = read(), y = read(), w = read();
// scanf("%d %d %d", &x, &y, &w);
add(x, y, w);
}
Dijkstra();
return 0;
}
原文:https://www.cnblogs.com/lifehappy/p/12920476.html