MOOC PTA 08-图8 How Long Does It Take

http://pta.patest.cn/pta/test/18/exam/4/question/631

构建图的邻接矩阵，寻找入度为0的顶点，将其压入队列，出队列时对其相连接的顶点入度减1，更新每个顶点的最大时间。

刚开始提交 3和5 测试点过不去 是因为我以为输出最后一个顶点就是工程的最后一个顶点

#include<vector>
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<queue>
using namespace std;
#define MAXnum 100
#define INF 100001
#define max(a,b) a>b?a:b
int Nv;//顶点个数
int Ne;//边的个数
int G[MAXnum][MAXnum];//图的邻接矩阵
int indegree[MAXnum];//入度计算
int time1[MAXnum];//时间计算
int countv;
void BuildGraph()
{
    int i,j,v1,v2,w;
    scanf("%d",&Nv);
    for(i=0; i<Nv; i++)
        for(j=0; j<Nv; j++)
            G[i][j]=-1;
    for( i=0; i<Nv; i++)
    {
        indegree[i]=0;
        time1[i]=0;
    }

    scanf("%d",&Ne);
    for(i=0; i<Ne; i++)
    {
        scanf("%d%d%d",&v1,&v2,&w);

        G[v1][v2]=w;
        indegree[v2]++;
        time1[v2]=max(w,time1[v2]);
    }

}
void topsort()
{
    queue<int> q;
    int v;
    int i;
    countv=0;
    
    for(i=0; i<Nv; i++)
        if(!indegree[i])
            q.push(i);


    while(!q.empty())
    {
        v=q.front();
        countv++;
        q.pop();
        
        for( i=0; i<Nv; i++)
            if(G[v][i]!=-1&&indegree[i]>0)
            {
                if(--indegree[i]==0)
                    q.push(i);
                    
                time1[i]=max(G[v][i]+time1[v],time1[i]);
            }
    }
    
    if(countv<Nv)
        printf("Impossible\n");
    else
    {
        int max1=-1;
        for(i=0; i<Nv; i++)
            if(max1<time1[i])
                max1=time1[i];
        printf("%d\n",max1);
    }
}
int main()
{
    BuildGraph();
    topsort();
    return 0;
}