題目大意：給無向簡單連通圖的m條邊，求從頂點1到頂點n的最短路，同時使得最短路徑中標記為0的邊數與不在最短路徑中標記為1的邊數之和最小，輸出這些邊。

#include<bits/stdc++.h>
using namespace std;
#define maxn 100005
struct Edge
{
    int x,y,z;
}edge[maxn<<1];

struct Node
{
    int v,dis,num,pre;
}node[maxn];

struct cmp
{
    bool operator() (Node &x,Node &y)
    {
        return x.dis>y.dis||(x.dis==y.dis&&x.num>y.num);
    }
};
vector<int>g[maxn];
bool vis[maxn];
bool used[maxn<<1];
int n,m;
priority_queue<Node,vector<Node>,cmp> Q;

void solve()
{
    Node temp=Node{1,1,0,-1};
    int i;
    Q.push(temp);
    memset(vis,0,sizeof(vis));
    while(!Q.empty())
    {
        temp=Q.top();Q.pop();
        if(vis[temp.v]) continue;
        vis[temp.v]=1;
        node[temp.v]=temp;
        if(temp.v==n) break;
        for(i=0;i<g[temp.v].size();++i)
        {
            int j=g[temp.v][i];
            if(vis[edge[j].y]) continue;
            if(edge[j].z) Q.push(Node{edge[j].y,temp.dis+1,temp.num,j});
            else Q.push(Node{edge[j].y,temp.dis+1,temp.num+1,j});
        }
    }

    int ans=0;
    memset(used,0,sizeof(used));
    i=temp.pre;
    while(i!=-1)
    {
        used[i]=1;
        i=node[edge[i].x].pre;
    }
    for(i=1;i<=m+m;i+=2)
    {
        if(used[i]||used[i+1])
        {
            if(edge[i].z==0) 
                ++ans;
        }
        else if(edge[i].z==1) 
            ++ans;
    }
    printf("%d\n",ans);
    for(i=1;i<=m+m;i+=2)
    {
        if(used[i]||used[i+1])
        {
            if(edge[i].z==0)  printf("%d %d 1\n",edge[i].x,edge[i].y);
        }
        else if(edge[i].z==1) printf("%d %d 0\n",edge[i].x,edge[i].y);
    }
}

int main()
{
    int i;
    cin>>n>>m;
    for(i=1;i<=m+m;i+=2)
    {
        scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].z);
        edge[i+1].x=edge[i].y;
        edge[i+1].y=edge[i].x;
        edge[i+1].z=edge[i].z;
        g[edge[i].x].push_back(i);
        g[edge[i+1].x].push_back(i+1);
    }
    solve();
    return 0;
}