Dijkstra's algorithm: seeking single source shortest path is the right in FIG.
1, the optimized code heap FIG Dijkstra:
1 #define INF 0x7fffffff 2 #include<queue> 3 #include<cstdio> 4 #include<iostream> 5 #include<algorithm> 6 using namespace std; 7 const int N = 100005; 8 const int M = 500005; 9 struct edge 10 { 11 int to,dis,next; 12 }e[M]; 13 int head[N],dis[N],cnt; 14 bool vis[N]; 15 int n,m,s; 16 void add(int u,int v,int d) 17 { 18 ++cnt; 19 e[cnt].dis=d; 20 e[cnt].to=v; 21 e[cnt].next=head[u]; 22 head[u]=cnt; 23 } 24 struct node 25 { 26 int dis,pos; 27 }; 28 bool operator < (const node &x,const node &y) 29 { 30 return x.dis<y.dis; 31 } 32 priority_queue<node> q; 33 inline void dijkstra() 34 { 35 dis[s]=0; 36 q.push((node){0,s}); 37 while(!q.empty()) 38 { 39 node tmp=q.top(); 40 q.pop(); 41 int x=tmp.pos,d=tmp.dis; 42 if(vis[x]) continue; 43 vis[x]=1; 44 for(int i=head[x];i;i=e[i].next) 45 { 46 int y=e[i].to; 47 if(dis[y]>dis[x]+e[i].dis) 48 { 49 dis[y]=dis[x]+e[i].dis; 50 if(!vis[y]) q.push((node){dis[y],y}); 51 } 52 } 53 } 54 } 55 int main() 56 { 57 cin>>n>>m>>s; 58 for(int i=1;i<=n;++i) dis[i]=INF; 59 for(int i=0;i<m;++i) 60 { 61 int u,v,d; 62 cin>>u>>v>>d; 63 add(u,v,d); 64 } 65 dijkstra(); 66 for(int i=1;i<=n;++i) 67 cout<<dis[i]<<" "; 68 cout<<endl; 69 return 0; 70 }
2, no stack optimized code to FIG Dijkstra:
1 #include<iostream> 2 #include<cstdio> 3 #include<vector> 4 #include<queue> 5 using namespace std; 6 const int N = 10005; 7 const int INF = 0x7fffffff; 8 struct node 9 { 10 int x,d; 11 node(int a,int b){x=a;d=b;} 12 }; 13 bool operator <(const node &x,const node &y) 14 { 15 return x.d<y.d; 16 } 17 vector<node> eg[N]; 18 int dis[N],n; 19 void Dijkstra(int s) 20 { 21 int i; 22 for(i=0;i<=n;i++) dis[i]=INF; 23 dis[s]=0; 24 priority_queue<node> q; 25 q.push(node(s,dis[s])); 26 while(!q.empty()) 27 { 28 node x=q.top();q.pop(); 29 for(i=0;i<eg[x.x].size();i++) 30 { 31 node y=eg[x.x][i]; 32 if(dis[y.x]>x.d+y.d) 33 { 34 dis[y.x]=x.d+y.d; 35 q.push(node(y.x,dis[y.x])); 36 } 37 } 38 } 39 } 40 int main() 41 { 42 int a,b,d,m; 43 cin>>n>>m>>s; 44 //处理多组数据时，使用 for(int i=0;i<n;++i) eg[i].clear(); 45 while(m--) 46 { 47 cin>>a>>b>>d; 48 eg[a].push_back(node(b,d)); 49 eg[b].push_back(node(a,d)); 50 } 51 Dijkstra(s); 52 cout<<dis[n]<<endl; 53 return 0; 54 }
3, entitled: Luogu shortest template 1 Luogu shortest template 2 Luogu shortest template 3
* When looking solution to a problem and found a big brother to use zkw Dijkstra's algorithm to optimize the tree line , where the code and paste the link:
1 #include <algorithm> 2 #include <cmath> 3 #include <cstdio> 4 #include <cstdlib> 5 #include <cstring> 6 #include <ctime> 7 #include <iostream> 8 #include <map> 9 #include <queue> 10 #include <set> 11 #include <stack> 12 #include <string> 13 #include <vector> 14 using namespace std; 15 #define go(i, j, n, k) for (int i = j; i <= n; i += k) 16 #define fo(i, j, n, k) for (int i = j; i >= n; i -= k) 17 #define rep(i, x) for (int i = h[x]; i; i = e[i].nxt) 18 #define mn 100010 19 #define mm 200020 20 #define inf 2147483647 21 #define ll long long 22 #define ld long double 23 #define fi first 24 #define se second 25 #define root 1, n, 1 26 #define lson l, m, rt << 1 27 #define rson m + 1, r, rt << 1 | 1 28 #define bson l, r, rt 29 inline int read(){ 30 int f = 1, x = 0;char ch = getchar(); 31 while (ch > '9' || ch < '0'){if (ch == '-')f = -f;ch = getchar();} 32 while (ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();} 33 return x * f; 34 } 35 inline void write(int x){ 36 if (x < 0)putchar('-'),x = -x; 37 if (x > 9)write(x / 10); 38 putchar(x % 10 + '0'); 39 } 40 //This is AC head above... 41 struct node{ 42 int v, nxt, w; 43 } e[mm << 1]; 44 int h[mn], p; 45 inline void add(int a,int b,int c){ 46 e[++p].nxt = h[a]; 47 h[a] = p; 48 e[p].v = b; 49 e[p].w = c; 50 } 51 int dis[mn]; 52 int n, m, s, t; 53 struct tree{ 54 int minw, minv; 55 }; 56 is struct SegmentTree { 57 is Tree Z [Mn << 2 ]; 58 inline void Update ( int RT) { 59 Z [RT] .minw min = (Z [RT << . 1 ] .minw, Z [RT << . 1 | . 1 ] .minw); // maintain minimum interval 60 Z [RT] .minv = (Z [RT << . 1 ] .minw <Z [RT << . 1 | . 1 ?] .minw) Z [RT < < . 1 ] .minv: Z [RT << . 1 | . 1 ] .minv; // maintain the position of the minimum interval 61 } 62 inline void Build ( int L, int R & lt, int RT) { // Contribution 63 is IF (L == R & lt) { 64 Z [RT] .minw == S = L? 0 ; INF: // We direct contribution the point s is set to 0 65 Z [RT] = L .minv; // record the minimum position, to facilitate the modification 66 return ; 67 } 68 int m = (L + R & lt) >> . 1 ; 69 Build (LSON); 70 Build (rson); 71 is Update (RT); 72 } 73 inline void modify(int l,int r,int rt,int now,int v){//单点修改 74 if(l==r){ 75 z[rt].minw = v; 76 return; 77 } 78 int m = (l + r) >> 1; 79 if(now<=m) 80 modify(lson, now, v); 81 else 82 modify(rson, now, v); 83 Update (RT); 84 } 85 } TR; 86 inline void Dij of () { // the core of Dijkstra's 87 Go (I, . 1 , n-, . 1 ) { 88 dis [I] = INF; 89 } // initialize dis 90 DIS [S] = 0 ; 91 is the while (tr.z [ . 1 ] .minw <INF) { // this is determined whether the air 92 int X = tr.z [ . 1 ] .minv; // get the entire segment tree smallest point 93 tr.modify (the root, X, INF); //Single point to modify the minimum point INF 94 REP (I, X) { 95 int V = E [I] .v; 96 IF (DIS [V]> DIS [X] + E [I] .W) { 97 DIS [V] = DIS [X] + E [I] .W; 98 tr.modify (the root, V, DIS [X] + E [I] .W); // this is similar to enqueue operation 99 } 100 } 101 } 102 } 103 int main () { 104 n-= Read (), m = Read (), S = Read (), T = Read (); 105 Go (I, . 1 , m, . 1 ) { 106 intRead = X (), Y = Read (), V = Read (); 107 the Add (X, Y, V); 108 the Add (Y, X, V); // this must remember not to be positive in FIG. anti two sides 109 } 110 tr.build (the root); // achievements 111 Dij of (); // the Dijkstra 112 COUT << DIS [T]; 113 return 0 ; 114 }
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