HDU 4109 Instrction Arrangement（DAG上的最长路）

把点编号改成1-N，加一点0，从0点到之前任意入度为0的点之间连一条边权为0的边，求0点到所有点的最长路。

SPFA模板留底用

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <queue>

#include <string>

#include <iostream>

#include <cmath>

#define ll long long

#define inf 1000000000

#define maxn 1010

using namespace std;



struct Edge

{

    int from,to;

    int dist;

    Edge(int a,int b,int c)

    {

        from=a,to=b,dist=c;

    }

};

struct SPFA

{

    int n,m;

    vector<Edge> edges;

    vector<int> g[maxn];

    int d[maxn];

    int p[maxn];

    int cnt[maxn];

    bool inq[maxn];



    void init(int n_)

    {

        n=n_;

        int i;

        for(i=0; i<=n; i++)

        {

            g[i].clear();

        }

        edges.clear();

    }



    void addedge(int from,int to,double dist)

    {

        edges.push_back(Edge(from,to,dist));

        m=edges.size();

        g[from].push_back(m-1);

    }



    void spfa(int s)

    {

        int i;

        queue<int> q;

        memset(inq,0,sizeof(inq));

        memset(cnt,0,sizeof(cnt));

        for(i=0; i<=n; i++)

        {

            d[i]=-inf;

        }

        d[s]=0;

        inq[s]=true;

        q.push(s);

        while(!q.empty())

        {

            int u=q.front();

            q.pop();

            inq[u]=false;

            for(i=0; i<g[u].size(); i++)

            {

                Edge& e=edges[g[u][i]];

                if(d[e.to]<d[u]+e.dist)

                {

                    d[e.to]=d[u]+e.dist;

                    p[e.to]=g[u][i];

                    if(!inq[e.to])

                    {

                        q.push(e.to);

                        inq[e.to]=true;

                    }

                }

            }

        }

        int ans=0;

        for(i=1;i<=n;i++)

        {

            ans=max(ans,d[i]);

        }

        printf("%d\n",ans+1);

    }

};

SPFA solver;

int N,M;

int ru[maxn];



int main()

{

   // freopen("input.txt","r",stdin);

    while ( scanf( "%d%d", &N, &M ) == 2 )

    {

        memset( ru, 0, sizeof(ru) );

        solver.init(N);

        for ( int i = 0; i < M; ++i )

        {

            int u, v, w;

            scanf( "%d%d%d", &u, &v, &w );

            u++,v++;

            ++ru[v];

            solver.addedge(u, v, w );

        }

        for ( int i = 1; i <= N; ++i )

        {

            if ( ru[i] == 0 )

                solver.addedge( 0, i, 0 );

        }

        solver.spfa(0);

    }

    return 0;

}