dijkstra+priority_queue+vector

最短路

时间限制: 3 Sec  内存限制: 128 MB

题目描述

给定M条边，N个点的带权无向图 
求1到N的最短路 
N<=100000 
M<=500000

输入

第一行：N,M 
接下来M行3个正整数：ai,bi,ci 表示ai,bi之间有一条长度为ci的路 
0<=ci<=1000

输出

一个整数，表示1到N的最短距离

样例输入

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

样例输出

2

提示

注意图中可能有重边和自环,数据保证1到N有路径相连

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<cstdlib>
 7 #include<vector>
 8 #include <queue>
 9 using namespace std;
10 typedef long long ll;
11 typedef long double ld;
12 typedef pair<int,int> pr;
13 const double pi=acos(-1);
14 #define rep(i,a,n) for(int i=a;i<=n;i++)
15 #define per(i,n,a) for(int i=n;i>=a;i--)
16 #define Rep(i,u) for(int i=head[u];i;i=Next[i])
17 #define clr(a) memset(a,0,sizeof(a))
18 #define pb push_back
19 #define mp make_pair
20 #define fi first
21 #define sc second
22 #define pq priority_queue
23 #define pqb priority_queue <int, vector<int>, less<int> >
24 #define pqs priority_queue <pr, vector<pr>, greater<pr> >
25 ld eps=1e-9;
26 ll pp=1000000007;
27 ll mo(ll a,ll pp){if(a>=0 && a<pp)return a;a%=pp;if(a<0)a+=pp;return a;}
28 ll powmod(ll a,ll b,ll pp){ll ans=1;for(;b;b>>=1,a=mo(a*a,pp))if(b&1)ans=mo(ans*a,pp);return ans;}
29 void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
30 //void add(int x,int y,int z){ v[++e]=y; next[e]=head[x]; head[x]=e; cost[e]=z; }
31 int dx[5]={0,-1,1,0,0},dy[5]={0,0,0,-1,1};
32 ll read(){ ll ans=0; char last=‘ ‘,ch=getchar();
33 while(ch<‘0‘ || ch>‘9‘)last=ch,ch=getchar();
34 while(ch>=‘0‘ && ch<=‘9‘)ans=ans*10+ch-‘0‘,ch=getchar();
35 if(last==‘-‘)ans=-ans; return ans;
36 }
37 #define N 100005
38 const int inf=1<<29;
39 vector<pr> vec[N];
40 pqs Q;
41 int dis[N],vis[N];
42 int main()
43 {
44     int n=read(),m=read();
45     for (int i=1;i<=m;i++) {
46         int a=read(),b=read(),c=read();
47         vec[a].pb(mp(b,c));
48         vec[b].pb(mp(a,c));
49     }
50     for (int i=1;i<=n;i++) dis[i]=inf;
51     dis[1]=0; Q.push(mp(dis[1],1));
52     while (!Q.empty()){
53         pr p=Q.top(); Q.pop(); int u=p.sc;
54         if (vis[u]) continue; vis[u]=1;
55         for (int i=0;i<vec[u].size();i++){
56             if (vec[u][i].sc+dis[u]<dis[vec[u][i].fi])
57                 dis[vec[u][i].fi]=vec[u][i].sc+dis[u];
58                 Q.push(mp(dis[vec[u][i].fi],vec[u][i].fi));
59         }
60     }
61     printf("%d",dis[n]);
62     return 0;
63  }