https://codeforces.com/contest/1473/problem/E
vector存圖：
1 #define bug(x) cout<<#x<<" is "<<x<<endl 2 #define IO std::ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); 3 #include <bits/stdc++.h> 4 #define iter ::iterator 5 using namespace std; 6 typedef long long ll; 7 typedef pair<int,ll>P; 8 #define pb push_back 9 #define mk make_pair 10 #define se second 11 #define fi first 12 #define rs o*2+1 13 #define ls o*2 14 const ll inf=1e18; 15 const int N=8e5+5; 16 17 int n,m; 18 19 vector<P>g[N]; 20 21 void add(int x,int y,ll z){ 22 g[x].pb(mk(y,z)); 23 g[y].pb(mk(x,z)); 24 25 g[x+n].pb(mk(y+n,z)); 26 g[y+n].pb(mk(x+n,z)); 27 28 g[x+2*n].pb(mk(y+2*n,z)); 29 g[y+2*n].pb(mk(x+2*n,z)); 30 31 g[x+3*n].pb(mk(y+3*n,z)); 32 g[y+3*n].pb(mk(x+3*n,z)); 33 34 g[x].pb(mk(y+n,0)); 35 g[y].pb(mk(x+n,0)); 36 37 g[x+2*n].pb(mk(y+3*n,0)); 38 g[y+2*n].pb(mk(x+3*n,0)); 39 40 g[x].pb(mk(y+2*n,2*z)); 41 g[y].pb(mk(x+2*n,2*z)); 42 43 g[x+n].pb(mk(y+3*n,2*z)); 44 g[y+n].pb(mk(x+3*n,2*z)); 45 46 g[x].pb(mk(y+3*n,z)); 47 g[y].pb(mk(x+3*n,z)); 48 } 49 50 ll dis[N]; 51 52 struct node{ 53 int u; 54 ll w; 55 bool operator <(const node& t)const{ 56 return w>t.w; 57 } 58 }; 59 60 void dij(){ 61 priority_queue<node>q; 62 fill(dis+1,dis+4*n+1,inf); 63 dis[1]=0; 64 q.push(node{1,0}); 65 while(!q.empty()){ 66 node now=q.top(); 67 q.pop(); 68 int u=now.u; 69 if(now.w!=dis[u])continue;//優化，如果已經更新了與u相連的點通過u點到起點的距離那麼無需再更新 70 for(int i=0;i<g[u].size();i++){ 71 int v=g[u][i].fi; 72 ll w=g[u][i].se; 73 //bug(w);a 74 if(dis[u]+w<dis[v]){ 75 dis[v]=dis[u]+w; 76 q.push(node{v,dis[v]}); 77 } 78 } 79 } 80 81 } 82 83 struct node1{ 84 int x,y; 85 ll z; 86 }e[N]; 87 88 int main(){ 89 IO; 90 cin>>n>>m; 91 for(int i=1;i<=m;i++){ 92 int u,v; 93 ll w; 94 cin>>u>>v>>w; 95 add(u,v,w); 96 } 97 dij(); 98 for(int i=2;i<=n;i++){ 99 //printf("%lld\n",dis[i+3*n]); 100 cout<<dis[i+3*n]<<" "; 101 } 102 cout<<endl; 103 }
前向星存圖
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 typedef long long LL; 5 const LL INF = 1e18; 6 const int N = 2e5 + 20; 7 8 struct Edge 9 { 10 int to, nxt, w; 11 }line[N * 20]; 12 int fist[N * 4], idx; 13 int n, m; 14 15 void add(int x, int y, int z) 16 { 17 line[idx] = (Edge){y, fist[x], z}; 18 fist[x] = idx ++; 19 } 20 21 void addedge(int x, int y, int z) 22 { 23 add(x, y, z); 24 25 add(x + n, y + n, z); 26 27 add(x + 2 * n, y + 2 * n, z); 28 29 dd(x + 3 * n, y + 3 * n, z); 30 31 add(x, y + n, 0); 32 33 add(x + 2 * n, y + 3 * n, 0); 34 35 add(x, y + 2 * n, 2 * z); 36 37 add(x + n, y + 3 * n, 2 * z); 38 39 add(x, y + 3 * n, z); 40 } 41 42 bool st[N * 4]; 43 LL dis[N * 4]; 44 struct zt 45 { 46 int x; 47 LL d; 48 }; 49 bool operator < (zt a, zt b) 50 { 51 return a.d > b.d; 52 } 53 54 void heap_dijkstra() 55 { 56 priority_queue<zt> q; 57 for(int i = 1; i <= 4 * n; ++ i) dis[i] = INF; 58 dis[1] = 0; 59 q.push((zt){1, 0}); 60 while(!q.empty()) 61 { 62 zt u = q.top(); q.pop(); 63 if(st[u.x]) continue; 64 st[u.x] = 1; 65 for(int i = fist[u.x]; i != -1; i = line[i].nxt) 66 { 67 int v = line[i].to; 68 if(dis[v] > dis[u.x] + line[i].w) 69 { 70 dis[v] = dis[u.x] + line[i].w; 71 q.push((zt){v, dis[v]}); 72 } 73 } 74 } 75 } 76 77 int main() 78 { 79 // freopen("E.in", "r", stdin); 80 memset(fist, -1, sizeof fist); 81 scanf("%d%d", &n, &m); 82 for(int i = 1; i <= m; ++ i) 83 { 84 int a, b, c; 85 scanf("%d%d%d", &a, &b, &c); 86 addedge(a, b, c); 87 addedge(b, a, c); 88 } 89 heap_dijkstra(); 90 for(int i = 2; i <= n; ++ i) 91 printf("%lld\n", dis[i]); 92 //puts(""); 93 return 0; 94 }