EZDIJKST – Easy Dijkstra Problem

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Determine the shortest path between the specified vertices in the graph given in the input data.
Hint: You can use Dijkstra’s algorithm.
Hint 2: if you’re a lazy C++ programmer, you can use set and cin/cout (with sync_with_stdio(0)) – it should suffice.

Input

first line – one integer – number of test cases
For each test case the numbers V, K (number of vertices, number of edges) are given,
Then K lines follow, each containing the following numbers separated by a single space: 
ai, bi, ci
It means that the graph being described contains an edge from ai to bi,
with a weight of ci.
Below the graph description a line containing a pair of integers A, B is present.
The goal is to find the shortest path from vertex A to vertex B.
All numbers in the input data are integers in the range 0..10000.

Output

For each test case your program should output (in a separate line) a single number C – the length of the shortest path from vertex A to vertex B. In case there is no such path, your program should output a single word “NO” (without quotes)

Example

Input:
3
3 2
1 2 5
2 3 7
1 3
3 3
1 2 4
1 3 7
2 3 1
1 3
3 1
1 2 4
1 3

Output:
12
5
NO

巩固一下dij算法，WA了好几次，原来是有向图。。。看题一定要认真T^T。愈发感觉邻接矩阵版太垃圾了，堆优化大法好

AC代码：

#include<iostream>
#include<string.h>
#include<algorithm>
#include<queue>  
using namespace std;  
typedef long long LL;    
    
const int MAXN = 300005;    
#define INF 0x3f3f3f3f  
  
vector<int> e[MAXN];//点   
vector<int> w[MAXN];//权   
bool vis[MAXN];  
LL dis[MAXN];//距离   
int pre[MAXN];//记录路径   
int n,m;//点数和边数   
struct node  
{  
    LL d;//储存距离   
    int u;//点的标号   
    bool operator<(const node & rhs) const  
    {  
        return d>rhs.d;//距离短的优先   
    }  
};  
  
  
  
int dij(int start)  
{  
    priority_queue<node> q;//优先队列   
    for(int i=1;i<=n;i++)  
    {  
        dis[i]=INF;//距离初始化   
    }  
    dis[start]=0;//出发点的距离设置为1   
    memset(vis,0,sizeof(vis));//初始化   
    node tn;  
    tn.d=0;//出发点的距离为0   
    tn.u=start;//出发点为1   
    q.push(tn);//入队列   
    while(!q.empty())//队列为空说明已经遍历完了地图   
    {  
        node t=q.top();//出队列（最短的出）   
        q.pop();  
        int u=t.u;  
        if(vis[u]) continue;//判断是否访问过   
        vis[u]=true;//访问后标记   
        for(int i=0;i<e[u].size();i++)  
        {  
             int v = e[u][i];   
             if(dis[v]>dis[u]+w[u][i])  
            {  
                dis[v]=dis[u]+w[u][i];//择优选择   
                pre[v]=u;//记录路径   
                tn.d=dis[v];//记录距离   
                tn.u=v;//设置跳板   
                q.push(tn);//跳板入队列   
            }     
        }  
          
    }  
}  
 int main()  
 {  
  int t;
  scanf("%d",&t);
  int cnt=1;
  while(t--)
  {
  	int start,end;
  	scanf("%d %d",&n,&m);
    for(int i=1;i<=n;i++)  
    {  
        e[i].clear();  
        w[i].clear();  
    }  
    int a,b,c;  
    for(int i=1;i<=m;i++)  
    {  
        scanf("%d %d %d",&a,&b,&c); 
        e[a].push_back(b);  
        w[a].push_back(c);   //有向图
    }  
    scanf("%d %d",&start,&end);
    dij(start);  
    if(dis[end]==INF) printf("NO\n");
	else printf("%lld\n",dis[end]);
}
    return 0;  
 }