Codeforces Round #257 (Div. 2) D题：Jzzhu and Cities 删特殊边的最短路_html/css

Codeforces Round #257 (Div. 2) D题：Jzzhu and Cities 删特殊边的最短路_html/css_WEB-ITnose

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发布： 2016-06-24 12:01:20

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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 mroads connecting the cities. One can go from city ui to vi (and vise versa) using the i-th road, the length of this road is xi. 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 si (and vise versa), the length of this route is yi.

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?≤?105; 1?≤?m?≤?3·105; 1?≤?k?≤?105).

Each of the next m lines contains three integers ui,?vi,?xi (1?≤?ui,?vi?≤?n; ui?≠?vi; 1?≤?xi?≤?109).

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Each of the next k lines contains two integers si and yi (2?≤?si?≤?n; 1?≤?yi?≤?109).

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.

Sample test(s)

input

5 5 31 2 12 3 21 3 33 4 41 5 53 54 55 5

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output

input

2 2 31 2 22 1 32 12 22 3

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output

思路：这题只要引入一个最短路条数，然后再遍历火车线，如果最短路与火车线长度相等，此时如果最短路条数是1的话，那说明这个最短路就是火车线，不能去掉，如果最短路条数大于1条，说明除了这条火车线外还有别的跟他同样短的路，可以去掉；如果火车线长度比最短路长的话，显然可以去掉。

这题刚开始看确实挺蛋疼的，比赛的时候也没啥想法。好久不做图论方面的了，所以有点不会了。

刚才敲spfa还重新看了下这算法才会手敲，确实有点生了。

刚开始queue没过，T了。然后改成优先队列就过了。呵呵。没明白。不是dijkstra才得用优先队列么，这题为什么也要用。后面想想，应该是先后问题T了。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<map>#include<queue>#include<set>#include<bitset>#define mem(a,b) memset(a,b,sizeof(a))#define INF 1000000070000using namespace std;typedef long long ll;typedef unsigned long long ull;int cnt,n,head[900005],vis[900005];int repeat[900005];//记录重复的边ll dist[900005],Map[900005];struct node{    int v,next,w;} e[900005];void add(int u,int v,int w){    e[cnt].v=v;    e[cnt].w=w;    e[cnt].next=head[u];    head[u]=cnt++;}void spfa(){    int i,j;    priority_queue<int>q;    for(i=0; i<=n; i++) dist[i]=INF;    repeat[1]=1;    q.push(1);    vis[1]=1;    dist[1]=0;    while(!q.empty())    {        int u=q.top();        q.pop();        vis[u]=0;        for(i=head[u]; i!=-1; i=e[i].next)        {            if(dist[e[i].v]>dist[u]+e[i].w)            {                dist[e[i].v]=dist[u]+e[i].w;                repeat[e[i].v]=repeat[u];                if(!vis[e[i].v])                {                    vis[e[i].v]=1;                    q.push(e[i].v);                }            }            else if(dist[e[i].v]==dist[u]+e[i].w)            {                repeat[e[i].v]+=repeat[u];                if(repeat[e[i].v]>=2) repeat[e[i].v]=2;            }        }    }}int main(){    int m,k,i,j,u,v,w,sum=0;//sum表示需要保留的火车线路    mem(head,-1);    cin>>n>>m>>k;    for(i=0;i<=n;i++)        Map[i]=INF;    for(i=0; i<m; i++)    {        scanf("%d%d%d",&u,&v,&w);        add(u,v,w);        add(v,u,w);    }    for(i=0; i<k; i++)    {        scanf("%d%d",&v,&w);        if(Map[v]>w) Map[v]=w;    }    for(i=1; i<=n; i++)    {        if(Map[i]!=INF)        {            add(1,i,Map[i]);            add(i,1,Map[i]);            sum++;        }    }    spfa();    for(i=1; i<=n; i++)        if(Map[i]!=INF)        {            if(dist[i]==Map[i]&&repeat[i]==2) sum--;            else if(dist[i]<Map[i]) sum--;        }    printf("%d\n",k-sum);//总的减去保留在最短路里面的    return 0;}

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