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
",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;
}