Codeforces 450D：Jzzhu and Cities（最短路，dijkstra）

D. Jzzhu and Cities

time limit per test: 2 seconds

memory limit per test: 256 megabytes

input: standard input

output: standard output

Jzzhu is the president of country A. There are \(n\) cities numbered from \(1\) to \(n\) in his country. City \(1\) is the capital of A. Also there are \(m\) roads connecting the cities. One can go from city \(u_i\) to \(v_i\) (and vise versa) using the \(i\)-th road, the length of this road is \(x_i\). Finally, there are \(k\) train routes in the country. One can use the \(i\)-th train route to go from capital of the country to city \(s_i\) (and vise versa), the length of this route is \(y_i\).

Jzzhu doesn't want to waste the money of the country, so he is going to close some of the train routes. Please tell Jzzhu the maximum number of the train routes which can be closed under the following condition: the length of the shortest path from every city to the capital mustn't change.

Input

The first line contains three integers \(n, m, k (2 ≤ n ≤ 10^5; 1 ≤ m ≤ 3\cdot10^5; 1 ≤ k ≤ 10^5)\).

Each of the next m lines contains three integers \(u_i, v_i, x_i (1 ≤ u_i, v_i≤ n; u_i ≠ v_i; 1 ≤ x_i ≤ 10^9)\).

Each of the next k lines contains two integers \(s_i\) and \(y_i\) \((2 ≤ s_i ≤ n; 1 ≤ y_i ≤ 10^9)\).

It is guaranteed that there is at least one way from every city to the capital. Note, that there can be multiple roads between two cities. Also, there can be multiple routes going to the same city from the capital.

Output

Output a single integer representing the maximum number of the train routes which can be closed.

Examples

input

5 5 3
1 2 1
2 3 2
1 3 3
3 4 4
1 5 5
3 5
4 5
5 5

output

2

input

2 2 3
1 2 2
2 1 3
2 1
2 2
2 3

output

2

題意

一個城市中有 \(m\) 條公路和 \(k\) 條鐵路，每條鐵路都和起點相連。如今要求在不改變起點到各點最短路徑長度的狀況下，拆除一些鐵路，問最多能夠拆除多少條鐵路

思路

將公路和鐵路放在一塊兒建圖，而後去跑最短路，在跑最短路的過程當中記錄一下每一個點的入度（該點被多少條最短路徑包含）

鐵路能夠刪除的條件：

1. 若是起點到該點的最短路徑和起點到該點的鐵路長度相等，判斷該點的入讀是否大於 \(1\)，若是大於 \(1\)，這條鐵路也是能夠刪除的（可以到達該點的最短路徑不止一條）
2. 起點到該點的最短路徑小於起點到該點的鐵路的長度

代碼

#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int maxn=1e6+10;
const int mod=1e9+7;
const int maxm=1e3+10;
using namespace std;
struct edge
{
    int to,Next;
    ll value;
}Edge[maxn];
int head[maxn];
int tot;
inline void add_edge(int u,int v,ll w)
{
    Edge[tot].to=v;
    Edge[tot].Next=head[u];
    Edge[tot].value=w;
    head[u]=tot++;
}
struct node
{
    int u;
    ll d;
    bool operator < (const node & dui) const{return d>dui.d;}
};
int ss[maxn];
ll yy[maxn];
ll dis[maxn];
int in[maxn];
inline void dijkstra(int s)
{
    priority_queue<node>que;
    que.push(node{s,0});
    dis[s]=0;
    while(!que.empty())
    {
        node res=que.top();
        que.pop();
        int u=res.u;ll d=res.d;
        if(d!=dis[u])
            continue;
        for(int i=head[u];~i;i=Edge[i].Next)
        {
            int v=Edge[i].to;
            ll w=Edge[i].value;
            if(dis[v]==dis[u]+w)
                in[v]++;
            if(dis[v]>dis[u]+w)
                in[v]=1,dis[v]=dis[u]+w,que.push(node{v,dis[v]});
        }
    }
}
int main(int argc, char const *argv[])
{
    #ifndef ONLINE_JUDGE
        freopen("/home/wzy/in", "r", stdin);
        freopen("/home/wzy/out", "w", stdout);
        srand((unsigned int)time(NULL));
    #endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,m,k;
    cin>>n>>m>>k;
    for(int i=0;i<=n;i++)
        dis[i]=INF;
    ms(head,-1);
    int x,y;
    ll z;
    while(m--)
        cin>>x>>y>>z,add_edge(x,y,z),add_edge(y,x,z);
    for(int i=0;i<k;i++)
        cin>>ss[i]>>yy[i],add_edge(1,ss[i],yy[i]),add_edge(ss[i],1,yy[i]);
    dijkstra(1);
    ll ans=0;
    for(int i=0;i<k;i++)
    {
        if(yy[i]==dis[ss[i]]&&in[ss[i]]>1)
            ans++,in[ss[i]]--;
        if(yy[i]>dis[ss[i]])
            ans++;
    }
    cout<<ans<<endl;
    #ifndef ONLINE_JUDGE
        cerr<<"Time elapsed:"<<1.0*clock()/CLOCKS_PER_SEC<<"s."<<endl;
    #endif
    return 0;
}