题目链接:点击进入
题目
思路
dijkstra
代码(python)
import heapq
# 定义无穷大
INF = float('inf')
# 初始化距离和访问数组
def init(n, s):
# 初始化所有节点的距离为无穷大
dis = [INF] * (n + 1)
# 源节点的距离为 0
dis[s] = 0
# 初始化所有节点未被访问
vis = [False] * (n + 1)
return dis, vis
# Dijkstra 算法实现
def dijkstra(n, s, graph):
dis, vis = init(n, s)
# 优先队列,存储 (距离, 节点)
pq = [(0, s)]
while pq:
# 取出当前距离最小的节点
dist, now = heapq.heappop(pq)
# 如果节点已经被访问过,跳过
if vis[now]:
continue
# 标记节点为已访问
vis[now] = True
# 遍历当前节点的所有邻接节点
for to, w in graph[now]:
# 如果通过当前节点到达邻接节点的距离更短
if dis[to] > dis[now] + w:
# 更新邻接节点的距离
dis[to] = dis[now] + w
# 将更新后的邻接节点加入优先队列
heapq.heappush(pq, (dis[to], to))
return dis
# 主函数
if __name__ == "__main__":
# 读取节点数、边数和源节点
n, m, s = map(int, input().split())
# 初始化图的邻接表
graph = [[] for _ in range(n + 1)]
# 读取每条边的信息
for _ in range(m):
x, y, w = map(int, input().split())
# 将边加入邻接表
graph[x].append((y, w))
# 调用 Dijkstra 算法求解最短路径
dis = dijkstra(n, s, graph)
# 输出每个节点到源节点的最短距离
for i in range(1, n + 1):
print(dis[i], end=" ")
代码(vector)
// Problem: P4779 【模板】单源最短路径(标准版)
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P4779
// Memory Limit: 125 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)
//#pragma GCC optimize(3)//O3
//#pragma GCC optimize(2)//O2
#include<iostream>
#include<string>
#include<map>
#include<set>
//#include<unordered_map>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include<stack>
#include<algorithm>
#include<iomanip>
#include<cmath>
#include<fstream>
#define X first
#define Y second
#define base 233
#define pb push_back
#define INF 0x3f3f3f3f3f3f3f3f
#define pii pair<int,int>
#define lowbit(x) x & -x
#define inf 0x3f3f3f3f
//#define int long long
//#define double long double
//#define rep(i,x,y) for(register int i = x; i <= y;++i)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double pai=acos(-1.0);
const int maxn=1e6+10;
const int mod=1e9+7;
const double eps=1e-9;
const int N=5e3+10;
/*--------------------------------------------*/
inline int read()
{
int k = 0, f = 1 ;
char c = getchar() ;
while(!isdigit(c)){if(c == '-') f = -1 ;c = getchar() ;}
while(isdigit(c)) k = (k << 1) + (k << 3) + c - 48 ,c = getchar() ;
return k * f ;
}
/*--------------------------------------------*/
int n,m,s;
int head[maxn],tot,dis[maxn];
bool vis[maxn];
struct node
{
int to;
int w;
int next;
}edge[maxn];
void init()
{
memset(head,-1,sizeof(head));
for(int i=0;i<=1e5;i++)
dis[i]=inf;
tot=0;
}
void add(int u,int v,int w)
{
edge[tot].to=v;
edge[tot].w=w;
edge[tot].next=head[u];
head[u]=tot++;
}
void dijkstra()
{
priority_queue<pii,vector<pii>,greater<pii>>q;
dis[s]=0;
q.push({dis[s],s});
while(q.size())
{
int now=q.top().second;
q.pop();
if(vis[now]) continue;
vis[now]=1;
for(int i=head[now];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(dis[v]>dis[now]+edge[i].w)
{
dis[v]=dis[now]+edge[i].w;
q.push({dis[v],v});
}
}
}
}
int main()
{
// ios::sync_with_stdio(false);
// cin.tie(0);cout.tie(0);
init();
cin>>n>>m>>s;
for(int i=1;i<=m;i++)
{
int x,y,w;
cin>>x>>y>>w;
add(x,y,w);
}
dijkstra();
for(int i=1;i<=n;i++)
cout<<dis[i]<<' ';
return 0;
}
代码(vector)
#include<iostream>
#include<algorithm>
#include<map>
#include<cmath>
#include<cstdio>
#include<queue>
#include<cstring>
#include<vector>
#define inf 0x3f3f3f3f
#define pii pair<int,int>
using namespace std;
typedef long long ll;
const int maxn=1e6+10;
int n,m,s,dis[maxn];
bool vis[maxn];
vector<pii>v[maxn];
void init()
{
memset(vis,0,sizeof(vis));
for(int i=0;i<100001;++i) dis[i]=inf;
}
void dijkstra()
{
priority_queue<pii,vector<pii>,greater<pii>>q;
dis[s]=0;
q.push({dis[s],s});
while(q.size())
{
int now=q.top().second;
q.pop();
if(vis[now]) continue;
vis[now]=1;
for(int i=0;i<v[now].size();++i)
{
int to=v[now][i].first;
int w=v[now][i].second;
if(dis[to]>dis[now]+w)
{
dis[to]=dis[now]+w;
q.push({dis[to],to});
}
}
}
}
int main()
{
init();
// scanf("%d%d%d",&n,&m,&s);
cin>>n>>m>>s;
for(int i=1;i<=m;++i)
{
int x,y,z;
// scanf("%d%d%d",&x,&y,&z);
cin>>x>>y>>z;
v[x].push_back({y,z});
}
dijkstra();
for(int i=1;i<=n;++i)
cout<<dis[i]<<" ";
// printf("%d ",dis[i]);
return 0;
}