我跑的是spfa
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<set>
#include<stack>
#include<vector>
#include<map>
#include<queue>
#define myself i,l,r
#define lson i<<1
#define rson i<<1|1
#define Lson i<<1,l,mid
#define Rson i<<1|1,mid+1,r
#define half (l+r)/2
#define inff 0x3f3f3f3f
#define lowbit(x) x&(-x)
#define PI 3.14159265358979323846
#define min4(a,b,c,d) min(min(a,b),min(c,d))
#define min3(x,y,z) min(min(x,y),min(y,z))
#define pii make_pair
#define pr pair<int,int>
const int dir[4][2]={0,1,0,-1,1,0,-1,0};
typedef long long ll;
const ll inFF=9223372036854775807;
typedef unsigned long long ull;
using namespace std;
const int maxn=1e5+5;
int head[maxn],sign;
int d[maxn],vis[maxn];
int n,m;
struct node
{
    int to,p,val;
}edge[maxn];
void add(int u,int v,int val)
{
    edge[sign]=node{v,head[u],val};
    head[u]=sign++;
}
void init()
{
    sign=0;
    memset(head,-1,sizeof(head));
}
int spfa()
{
    for(int i=0;i<=n;i++)
    {
        d[i]=0;
        vis[i]=0;
    }
    d[1]=inff;
    queue<int> q;
    q.push(1);
    while(!q.empty())
    {
        int u=q.front();
        q.pop(),vis[u]=0;
        for(int i=head[u];~i;i=edge[i].p)
        {
            int v=edge[i].to;
            if(d[v]<min(edge[i].val,d[u]))
            {
                d[v]=min(edge[i].val,d[u]);
                if(!vis[v])
                {
                    q.push(v);
                    vis[v]=1;
                }
            }
        }
    }
    return d[n];
}
int main()
{
    int t,x,y,z,p=0;
    cin>>t;
    while(t--)
    {
        scanf("%d %d",&n,&m);
        init();
        for(int i=1;i<=m;i++)
        {
            scanf("%d %d %d",&x,&y,&z);
            add(x,y,z),add(y,x,z);
        }
        printf("Scenario #%d:\n",++p);
        cout<<spfa()<<endl;
        printf("\n");
    }
    return 0;
}

• Re:提供一种思路：可以吧1点的值设为无穷，然后其他点赋值为0，应该很多算法都可以跑 (2158) LittleMelon6 2022-11-26 20:48:56