如何在priority_queue< PI>向量< PI>中减小针对特定边缘的密钥. ，更大的PI. &gt ;，尝试实施prim的算法? [英] How to decrease key for a particular edge in a priority_queue<PI, vector<PI> ,greater<PI> >,trying to implement prim's algorithm?

问题描述

#include <bits/stdc++.h>
 using namespace std;

#define fast ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define LL long long int
#define pb push_back
#define mp make_pair
#define PI pair<int,int>
#define PL pair<LL,LL>
#define PIS pair< int,string>


#define test int t;cin>>t;while(t--)
#define ff first
#define ss second
#define INF 1000000000
#define input(a,n) for(i=1;i<=n;i++)cin>>a[i];
#define output(a,n) for(i=1;i<=n;i++)cout<<a[i]<<" ";
vector< vector<LL> >v(3002, vector<LL>(3002,-1));
priority_queue<PI, vector<PI> ,greater<PI> > pp;
LL w=0;
int vis[3002]={0};
/*void deck(int a ,int b, int *k)
{
    while(!pp.empty())
    {
        i=pp.top.ss;
        if(i==a)
          pp.top
    }
}*/
void prim(LL s, LL *k, LL *p,LL n)
{

    pp.push(mp(0,s));
    k[s]=0;
    LL i,x,a,b,c=0;
    vector<PI >::iterator it;
    while(true)
    {
        if(c==n)
          break;
        i=pp.top().ss;
        //cout<<i<<" ";
        if(vis[i]!=1)
        w=w+pp.top().ff;
        vis[i]=1;
        c++;
        pp.pop();
        for(x=1;x<=n;x++)
        {
            if(v[i][x]!=-1)
            {

            a=x;
            b=v[i][x];
            if(!vis[a] && b<k[a])
            {
                k[a]=b;
                p[a]=i;
                pp.push(mp(b,a));

            }
           }
       }
    }
}
int main()
{
    fast

    LL n,m,x,i,j,r,s;
    /*pp.push(mp(2,3));
    pp.push(mp(3,4));*/
    cin>>n>>m;
    LL k[n+1],p[n+1];
    v.resize(n+1);
    for(x=1;x<n+1;x++)
    {
        k[x]=INF;
        p[x]=-1;
    }
    for(x=0;x<m;x++)
    {
        cin>>i>>j>>r;
        /*v[i].pb(mp(j,r));
        v[j].pb(mp(i,r));*/
        if(v[i][j]!=-1)
        {
            if(v[i][j]>r)
            {
                v[i][j]=r;
                v[j][i]=r;
            }
       }
       else
       {
           v[i][j]=r;
           v[j][i]=r;
       }


    }
    cin>>s;
    prim(s,k,p,n);
    cout<<w;
    //cout<<pp.top().ss;    
}

我无法实现搜索特定内容的功能 值，即顶点并更改其键值，而不是我按下了 更改对，使用

I was not able to implement the function which searches a particular value i.e the vertex and changes it's key value instead I pushed the changed pair, using

pp.push(mp(b,a));

我可以通过使用if语句来获得一些测试用例

I was able to get some test cases right by using the if statement

if(c==n)
break;

其中"c"代表访问的顶点数.

where 'c' represents the count of vertexes visited.

推荐答案

我找到了一种使用以下方法实现算法的方法

I found a way to implement the algorithm using

std::priority_queue

通过更改方法，希望共享我的代码

By changing approach, would like to share my code

    #include<iostream>
    #include<algorithm>
    #include<vector>
    #include<set>
    #include<limits.h>
    #include<map>
    #include<stack>
    #include<stdio.h>
    #include<queue>
    #include<cmath>
    #include<string.h>

    using namespace std;
    #define pb push_back
    #define mp make_pair
    #define ff first
    #define ss second
    #define PII pair<int,int>
    vector<PII> v[3001];
    bool vis[3001];
    int sum=0;
    void prim(int s)
    {
        int y,x,i;
        PII p;
        priority_queue<PII, vector<PII> , greater<PII> > q;
        q.push(mp(0,s));
        while(!q.empty())
        {
            p = q.top();
            q.pop();
            x = p.ss;
            if(vis[x]==true)
                continue;
            sum+=p.ff;
            vis[x]=true;
            for(i=0;i<v[x].size();i++)
            {
                y = v[x][i].ss;
                if(vis[y]==false)
                    q.push(v[x][i]);
            }
        }
    }
        int main() {

        fast
        int f1=0,max=0,y,a1,x,j,w=0,f=0,l,m,b,c1=0,r=0;
        char t,a[4][4],c;
        int n,i=0,d=1,k;
        cin>>n>>k;
        for(x=0;x<k;x++)
        {
            cin>>i>>j>>r;
            v[i].pb(mp(r,j));
            v[j].pb(mp(r,i));
        }
        cin>>i;
        prim(i);
        cout<<sum;
        return 0;

}

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