图结构练习——最短路径

题目

坑点

题目输入的时候会有重边。比如说：

2 2
1 2 1
1 2 2

答案应该是1。

题解

就是jd最短路模板题，不多说

代码1：临界矩阵

使用临界矩阵，注意判断重边

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define PII pair<int, int>
#define x first
#define y second
const int maxn = 1e6 + 10;
const int INF = 0x3f3f3f3f;

const ll M = 11092019;
const int N = 100010;
int g[110][110];
int n, m;
int vis[110];
int dis[110];
void dj()
{
    priority_queue<PII, vector<PII>, greater<PII>> q;
    q.push({0, 1});
    dis[1] = 0;
    while (q.size())
    {
        PII k = q.top();
        q.pop();
        if (vis[k.second])
            continue;
        vis[k.second] = 1;
        for (int i = 1; i <= n; i++)
        {
            if (g[k.second][i])
            {
                int dist = k.first + g[k.second][i];
                if (dis[i] > dist)
                {
                    dis[i] = dist;
                    q.push({dis[i], i});
                    //cout<<dis[i]<<" "<<i<<endl;
                }
            }
        }
    }
}
int main()
{
    while (cin >> n >> m)
    {
        memset(g, 0, sizeof g);
        memset(dis, 0x3f, sizeof dis);
        memset(vis, 0, sizeof vis);
        for (int i = 0; i < m; i++)
        {
            int a, b, c;
            cin >> a >> b >> c;
            if (!g[a][b])//判断重边
                g[a][b] = g[b][a] = c;
            else
            {
                g[a][b] = g[b][a] = min(c, g[a][b]);
            }
        }
        dj();
        cout << dis[n] << endl;
    }

    return 0;
}

代码2:邻接表

邻接表就不需要判断重边，因为两个边都会输入

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define PII pair<int, int>
#define x first
#define y second
const int maxn = 1e6 + 10;
const int INF = 0x3f3f3f3f;

const ll M = 11092019;
const int N = 100010;
vector<PII> g[110];
int n, m;
int vis[110];
int dis[110];
void dj()
{
    priority_queue<PII, vector<PII>, greater<PII>> q;
    q.push({0, 1});
    dis[1] = 0;
    while (q.size())
    {
        PII k = q.top();
        q.pop();
        if (vis[k.second])
            continue;
        vis[k.second] = 1;
        for(PII t: g[k.second]){
            int dist=t.first+k.first;
            int ver=t.second;
            if(dist<dis[ver]){
                dis[ver]=dist;
            //    cout<< ver<< " "<<dis[ver]<<endl;
                q.push({dis[ver],ver});
            }
        }
    }
}
int main()
{
    while (cin >> n >> m)
    {
   for(int i=0;i<n;i++){
       g->clear();
   }
   memset(g, 0, sizeof g);
        memset(dis, 0x3f, sizeof dis);
        memset(vis, 0, sizeof vis);
        for (int i = 0; i < m; i++)
        {
            int a, b, c;
            cin >> a >> b >> c;
        g[a].push_back({c,b});
        g[b].push_back({c,a});
        }
        dj();
        cout << dis[n] << endl;
    }

    return 0;
}