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  • HDU 1874 畅通工程续(最短路训练

    因为数据比较小  

    所以flyod spfa dijkstra 多可以过

    Floyd 

    #include <bits/stdc++.h>
    using namespace std;
    const int INF = 0x3f3f3f3f;
    const int maxn = 210;
    
    int s[maxn][maxn];
    int main (){
        int n,m;
        while(~scanf("%d %d",&n,&m) ){
            for(int i=0;i<n;i++){
                for(int j=0;j<n;j++){
                    if(i != j)
                        s[i][j] = INF;
                    else
                        s[i][j] = 0;
                }
            }
            for(int i=0;i<m;i++){
                int x,y,v; scanf("%d %d %d",&x,&y,&v);
                s[x][y] = min(s[x][y],v);
                s[y][x] = min(s[y][x],v);
            }
            for(int k=0;k<n;k++)
                for(int i=0;i<n;i++)
                    for(int j=0;j<n;j++)
                        s[i][j] = min(s[i][j],s[i][k]+s[k][j]);
            int x,y;
            scanf("%d %d",&x,&y);
            if(s[x][y] == INF)
                printf("-1
    ");
            else
                printf("%d
    ",s[x][y]);
        }
    }


    spfa

    // Spfa 算法
    
    #include <bits/stdc++.h>
    using namespace std;
    const int maxn = 210;
    const int INF = 0x3f3f3f3f;
    
    vector <pair<int,int> >E[maxn];
    int d[maxn],inq[maxn];//inq 代表的是是否in queue
    int n,m;
    void init(){
        for(int i=0;i<maxn;i++) d[i] = INF;
        for(int i=0;i<maxn;i++) inq[i] = 0;
        for(int i=0;i<maxn;i++) E[i].clear();
    }
    int main()
    {
        while (~scanf("%d %d",&n,&m)){
            init();
            for(int i=1;i<=m;i++){
                int x,y,v;
                scanf("%d %d %d",&x,&y,&v);
                E[x].push_back(make_pair(y,v));
                E[y].push_back(make_pair(x,v));
            }
            queue<int>Q;
            int s,e;
            scanf("%d %d",&s,&e);
            d[s] = 0;
            inq[s] = 1;//起点进队列
            Q.push( s );
            while (Q.size()){
                int now = Q.front();
                Q.pop(); inq[now] = 0;//该点出队列
                for(int i=0;i < E[now].size();i++){
                    int y = E[now][i].first;
                    int v = E[now][i].second;
                    if(d[y] > d[now] + v )
                    {
                        d[y] = d[now] + v;
                        if( inq[y] )//如果在队列
                            continue;
                        inq[y] = 1;
                        Q.push(y);
                    }
                }
            }
            if(d[e] == INF)
                printf("-1
    ");
            else
                printf("%d
    ",d[e]);
    
        }
        return 0;
    }

    dijkstra  //待测试

    #include <bits/stdc++.h>
    using namespace std;
    const int maxn = 210;
    const int INF = 0x3f3f3f3f;
    int n,m;
    vector<pair<int,int> > E[maxn];
    int d[maxn];
    void init(){
        for(int i=0;i<maxn;i++)
            E[i].clear() ,d[i] = INF;
    }
    int main()
    {
        while (~scanf("%d %d",&n,&m)){
            init();
            for(int i=1;i<=m;i++){
                int x,y,v;
                scanf("%d%d%d",&x,&y,&v);
                E[x].push_back(make_pair(y,v));
                E[y].push_back(make_pair(x,v));
            }
            int s,e;//start end;
            scanf("%d %d",&s,&e);
            d[s] = 0;
            int v = -1;
            priority_queue<pair<int,int> > Q;
            Q.push(make_pair(-d[s],s));
            while ( !Q.empty() )
            {
                int now = Q.top().second;
                if( now == e)//相等就可以直接跳出了   
                    break;
                Q.pop();
                for(int i=0;i < E[now].size();i++)
                {
                    v = E[now][i].first;
                    if(d[v] > d[now] + E[now][i].second )
                    {
                        d[v] = d[now] + E[now][i].second;
                        Q.push(make_pair(-d[v],v));
                    }
                }
    
            }
            if(d[e] == INF)
                printf("-1
    ");
            else
                printf("%d
    ",d[e]);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Draymonder/p/7337647.html
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