[kuangbin带你飞]专题四 最短路练习
作者：互联网
Til the Cows Come HomePOJ - 2387
题意：给你一幅地图，点1~N，双向正权路，问你N到1的最短路径？
算法：Dijkstra或者SPFA
思路：因为是正权路径，可以用Dijkstra算法；SPFA区别就是可以判断负权环。
1 #include <iostream>
2 #include <vector>
3 #include <cstring>
4 #include <algorithm>
5 #include <queue>
6 #include <cstdio>
7 #include <map>
8 #include <math.h>
9
10 using namespace std;
11
12 const int INF = 0x3f3f3f3f;
13 const int MAXN = 1000+10;
14
15 struct qnode{
16 int v,c;
17 qnode(int _v=0,int _c=0):v(_v),c(_c) {}
18 bool operator < (const qnode &r)const{
19 return c>r.c;
20 }
21 };
22 struct Edge{
23 int v,cost;
24 Edge(int _v=0,int _cost=0):v(_v),cost(_cost) {}
25 };
26 vector<Edge>E[MAXN];
27 void addedge(int u,int v,int w){E[u].push_back(Edge(v,w));}
28 bool vis[MAXN];
29 int dist[MAXN];
30 void Dijkstra(int n,int start){
31 memset(vis,false,sizeof vis);
32 for(int i=1;i<=n;i++) dist[i]=INF;
33 priority_queue<qnode> q;
34 dist[start]=0;
35 q.push(qnode(start,0));
36 while(!q.empty()){
37 qnode tmp = q.top();q.pop();
38 int u = tmp.v;
39 if(vis[u]) continue;
40 vis[u]=true;
41 for(int i=0;i<E[u].size();i++){
42 int v = E[u][i].v , cost = E[u][i].cost;
43 if(!vis[v]&&dist[v]>dist[u]+cost){
44 dist[v] = dist[u] + cost;
45 q.push(qnode(v,dist[v]));
46 }
47 }
48 }
49 }
50 int main()
51 {
52 int T,N;
53 cin>>T>>N;
54 while(T--){
55 int a,b,c;cin>>a>>b>>c;
56 addedge(a,b,c) , addedge(b,a,c);
57 }
58 Dijkstra(N,1);
59 cout<<dist[N]<<endl;
60 return 0;
61 }
Dijkstra
1 #include <iostream>
2 #include <vector>
3 #include <cstring>
4 #include <algorithm>
5 #include <queue>
6 #include <cstdio>
7 #include <map>
8 #include <math.h>
9
10 using namespace std;
11
12 const int INF = 0x3f3f3f3f;
13 const int MAXN = 1000+10;
14
15 struct Edge{
16 int v,cost;
17 Edge(int _v=0,int _cost=0):v(_v),cost(_cost) {}
18 };
19 vector<Edge> E[MAXN];
20 void addedge(int u,int v,int w){E[u].push_back(Edge(v,w));}
21 bool vis[MAXN];
22 int cnt[MAXN],dis[MAXN];
23 bool SPFA(int start,int n){
24 memset(vis,false,sizeof vis);
25 for(int i=1;i<=n;i++) dis[i] = INF;
26 vis[start]=true,dis[start]=0;
27 queue<int> q;q.push(start);
28 memset(cnt,0,sizeof cnt);cnt[start] = 1;
29 while(q.size()){
30 int u = q.front();q.pop();
31 vis[u]=false;
32 for(int i=0;i<E[u].size();i++){
33 int v = E[u][i].v;
34 if(dis[v]>dis[u]+E[u][i].cost){
35 dis[v]=dis[u]+E[u][i].cost;
36 if(!vis[v]){
37 vis[v] = true;
38 q.push(v);
39 if(++cnt[v]>n) return false;
40 }
41 }
42 }
43 }
44 return true;
45 }
46 int main()
47 {
48 int T,N;
49 cin>>T>>N;
50 while(T--){
51 int a,b,c;cin>>a>>b>>c;
52 addedge(a,b,c) , addedge(b,a,c);
53 }
54 SPFA(1,N);
55 cout<<dis[N]<<endl;
56 return 0;
57 }
SPFA
FroggerPOJ - 2253
题意：青蛙想要从石头1跳到石头2；其他石头是编号3~n；起点为石头1，终点石头2；问你们从起点到终点，跳到最大距离的最小可能是多少？
算法：SPFA
思路：这里只要改变路径改变方式就行了，dist[i]变成从起点到该点i途中，最大跳跃距离；
1 #include <iostream>
2 #include <vector>
3 #include <cstring>
4 #include <algorithm>
5 #include <queue>
6 #include <cstdio>
7 #include <map>
8 #include <math.h>
9 #include <iomanip>
10
11 using namespace std;
12
13 const int INF = 0x3f3f3f3f;
14 const int MAXN = 200+10;
15
16 struct node{
17 int x,y;
18 node(int a,int b):x(a),y(b) {}
19 node() {}
20 int d(int a,int b){
21 a = a - x , b = b - y;
22 return a*a + b*b;
23 }
24 }po[MAXN];
25 bool vis[MAXN];
26 int dis[MAXN];
27 bool SPFA(int start,int n){
28 memset(vis,false,sizeof vis);
29 for(int i=1;i<=n;i++) dis[i] = INF;
30 vis[start]=true,dis[start]=0;
31 queue<int> q;q.push(start);
32 while(q.size()){
33 int u = q.front();q.pop();
34 vis[u]=false;
35 for(int i=1;i<=n;i++){
36 if(i==u) continue;
37 int d = max( dis[u] , po[u].d(po[i].x , po[i].y));
38 if(dis[i]>d){
39 dis[i]=d;
40 if(!vis[i]){
41 vis[i] = true;
42 q.push(i);
43 }
44 }
45 }
46 }
47 return true;
48 }
49 int main()
50 {
51 int n,cas=1;
52 while(cin>>n&&n){
53 cout<<"Scenario #"<<cas++<<endl;
54 cout<<"Frog Distance = ";
55 for(int i=1;i<=n;i++) cin>>po[i].x>>po[i].y;
56 SPFA(1,n);
57 double r = dis[2];
58 cout<<fixed<<setprecision(3)<<sqrt(r)<<endl;
59 cout<<endl;
60 }
61 return 0;
62 }
View Code
Heavy Transportation POJ - 1797
题意：
标签：专题,const,int,短路,vis,cost,MAXN,kuangbin,include 来源： https://www.cnblogs.com/BugClearlove/p/16695116.html