很裸的最短路,不过节点数和边数都是1e6,直接dij肯定是不行了,稀疏图用heap优化一下就好
o(╯□╰)o注意STL里面的优先队列是优先级大的(值大的)在前面的,一开始没注意WA了好几发,哎,太粗心了
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <climits> #include <string> #include <iostream> #include <map> #include <cstdlib> #include <list> #include <set> #include <queue> #include <stack> using namespace std; typedef long long LL; const int maxn = 1000000 + 5; const LL INF = 10000000005LL; int u[maxn],v[maxn],w[maxn]; long long d[maxn]; int first[maxn],nxt[maxn]; bool vis[maxn]; int n,m; void add(int _u,int id) { nxt[id] = first[_u]; first[_u] = id; } struct Node { long long a; int b; Node(long long a,int b):a(a),b(b) {} bool operator < (const Node &x) const { return a > x.a; } }; //优先队列优化的dijkstra void dijkstra(int *v) { memset(vis,0,sizeof(vis)); for(int i = 1;i <= n;i++) d[i] = INF; d[1] = 0; priority_queue<Node> q; q.push(Node(d[1],1)); while(!q.empty()) { Node now = q.top(); q.pop(); int x = now.b; if(vis[x]) continue; vis[x] = true; for(int i = first[x];i != 0;i = nxt[i]) { if(d[v[i]] > d[x] + w[i]) { d[v[i]] = d[x] + w[i]; q.push(Node(d[v[i]],v[i])); } } } } int main() { int T; scanf("%d",&T); for(int kase = 1;kase <= T;kase++) { memset(first,0,sizeof(first)); memset(nxt,0,sizeof(nxt));; long long ans = 0; scanf("%d%d",&n,&m); for(int i = 1;i <= m;i++) { scanf("%d%d%d",&u[i],&v[i],&w[i]); add(u[i],i); } dijkstra(v); for(int i = 1;i <= n;i++) ans += d[i]; memset(first,0,sizeof(first)); memset(nxt,0,sizeof(nxt)); for(int i = 1;i <= m;i++) { add(v[i],i); } dijkstra(u); for(int i = 1;i <= n;i++) ans += d[i]; cout << ans << endl; } return 0; }