筆問題記録-2 最短ルート

テストサイト: グラフ理論 - 最短経路 - ダイクストラ


問題解決:
C++

#include <iostream>
#include <vector>
#include <queue>
using namespace std;
const long long inf = 0x3f3f3f3f3f3f3f3fLL;
const int num = 3e5+2;
struct edge
{
    
    
    int from,to;
    long long w;
    edge(int a,int b,long long c)
    {
    
    
        from = a;
        to = b;
        w = c;
    }
};


vector<edge> e[num];

struct s_node
{
    
    
    int id;
    long long n_dis;
    s_node(int b,long long c)
    {
    
    
        id = b;
        n_dis = c;
    }
    bool operator < (const s_node a) const
    {
    
    
        return n_dis > a.n_dis;
    }
};

int n,m;
long long dis[num];

void dijstra()
{
    
    
    int s = 1;
    bool done[num];
    for(int i = 1;i <= n;i++)
    {
    
    
        dis[i] = inf;
        done[i] = false;
    }
    dis[s] = 0;
    priority_queue <s_node> Q;
    Q.push(s_node(s,dis[s]));
    while(!Q.empty())
    {
    
    
        s_node u = Q.top();
        Q.pop();
        if(done[u.id]) continue;
        done[u.id] = true;
        
        for(int i = 0;i < e[u.id].size();i++)
        {
    
    
            edge y = e[u.id][i];
            if(done[y.to]) continue;
            if(dis[y.to] > y.w +u.n_dis)
            {
    
    
                dis[y.to] = y.w + u.n_dis;
                Q.push(s_node(y.to,dis[y.to]));
            }
        }
    }
}


int main()
{
    
    
    cin >> n >> m;
    for(int i = 1;i <= n;i++) e[i].clear();
    while(m--)
    {
    
    
        int u,v,w;
        cin >> u >> v >> w;
        e[u].push_back(edge(u,v,w));
    }
    
    dijstra();
    
    for(int i = 1;i <= n;i++)
    {
    
    
        if(dis[i] >= inf) cout << "-1" << " ";
        else cout << dis[i] << " ";
    }
    
    return 0;
}

パイソン

import heapq
import sys
n,m=map(int,input().split())
distance=[[] for i in range(n+1)]
for i in range(m):
  u,v,w=map(int,input().split())
  distance[u].append((v,w))
dp=[sys.maxsize]*(n+1)
q=[]
heapq.heappush(q,(0,1))
dp[1]=0
def dijkstra():
  visit=[0 for i in range(n+1)]
  while q:
    v=heapq.heappop(q)[1]
    if visit[v]==1:
      continue
    visit[v]=1
    for a,b in distance[v]:
      if visit[a]==1:
        continue
      dp[a]=min(dp[a],dp[v]+b)
      heapq.heappush(q,(dp[a],a))
dijkstra()
print(0,end=" ")
for i in range(2,n+1):
  if dp[i]==sys.maxsize:
    print(-1,end=" ")
  else:
    print(dp[i],end=" ")

他の偉人たちの問題解決のアイデアを調べましたが、それは非常に強力です

# 请在此输入您的代码
import math
import heapq


def djs(s):
    done = [0 for _ in range(n + 1)]  # bool矩阵
    hp = []
    dis[s] = 0
    heapq.heappush(hp, (0, s))  
    while hp:
        t = heapq.heappop(hp)[1]  # 取出该节点
        if done[t]:
            continue
        done[t] = 1
        for i in range(len(g[t])):  # 找t的邻居节点和距离
            v, w = g[t][i]
            if done[v]:
                continue
            if dis[v] > dis[t] + w:
                dis[v] = dis[t] + w
                heapq.heappush(hp, (dis[v], v))


n, m = map(int, input().split())
g = [[] for _ in range(n + 1)]  # 用邻接表来储存图
dis = [math.inf] * (n + 1)  
for i in range(m):
    u, v, w = map(int, input().split())
    g[u].append((v, w))  # 用邻接表来储存图
djs(1)
for i in range(1, n+1):
    if dis[i] >= math.inf:
        print(-1, end=' ')
    else:
        print(dis[i], end=' ')