最短路径
继续屯代码
floyed
1 #include<cstdio> 2 #include<cmath> 3 #include<cstring> 4 double map[101][101],x[101],y[101]; 5 int n,m,u,v; 6 double dis(int u,int v){ 7 return (sqrt((x[u]-x[v])*(x[u]-x[v])+(y[u]-y[v])*(y[u]-y[v]))); 8 } 9 10 11 int main(){ 12 scanf("%d",&n); 13 for (int i=1;i<=n;i++) scanf("%lf%lf",&x[i],&y[i]); 14 scanf("%d",&m); 15 for (int i=0;i<=n;i++) for (int j=0;j<=n;j++) map[i][j]=100000; 16 for (int i=0;i<m;i++){ 17 scanf("%d%d",&u,&v); 18 map[u][v]=dis(u,v); 19 map[v][u]=map[u][v]; 20 } 21 for (int k=1;k<=n;k++) 22 for (int i=1;i<=n;i++) 23 for (int j=1;j<=n;j++) if (map[i][k]+map[k][j]<map[i][j]) map[i][j]=map[i][k]+map[k][j]; 24 scanf("%d%d",&u,&v); 25 printf("%.2lf",map[u][v]); 26 }
迪杰斯塔拉
1 #include<cstring> 2 #include<cstdio> 3 #include<vector> 4 using namespace std; 5 const int MAXN=10000; 6 vector<int> W[MAXN],E[MAXN]; 7 int dis[MAXN],ans=0,n,m,map[MAXN][MAXN]; 8 bool vis[MAXN]={0}; 9 void add(int u,int v,int w1){ 10 E[u].push_back(v); 11 W[u].push_back(w1); 12 } 13 14 void dij(){ 15 memset(dis,127,sizeof(dis)); 16 dis[1]=0; 17 for (int i=0;i<n;i++){ 18 int min=100000,minn; 19 for (int j=1;j<=n;j++){ 20 if (dis[j]<min&&!vis[j]){ 21 min=dis[j]; 22 minn=j; 23 } 24 } 25 vis[minn]=1; 26 ans+=min; 27 for (int j=0;j<E[minn].size();j++){ 28 if (W[minn][j]+dis[minn]<dis[E[minn][j]]){ 29 dis[E[minn][j]]=W[minn][j]+dis[minn]; 30 } 31 } 32 } 33 } 34 35 36 int main(){ 37 scanf("%d",&n); 38 int u,v,w1; 39 for (int i=1;i<=n;i++){ 40 for (int j=1;j<=n;j++) scanf("%d",&map[i][j]); 41 } 42 43 for (int i=1;i<=n;i++) 44 for (int j=1;j<i;j++) { 45 add(j,i,map[i][j]); 46 add(i,j,map[i][j]); 47 } 48 dij(); 49 printf("%d",dis[n]); 50 }
优先队列优化后
1 #include<cstring> 2 #include<cstdio> 3 #include<vector> 4 #include<functional> 5 #include<queue> 6 using namespace std; 7 const int MAXN=10000; 8 vector<int> W[MAXN],E[MAXN]; 9 int dis[MAXN],ans=0,n,m,map[MAXN][MAXN]; 10 bool vis[MAXN]={0}; 11 struct dott{int num,v; 12 }; 13 void add(int u,int v,int w1){ 14 E[u].push_back(v); 15 W[u].push_back(w1); 16 } 17 18 struct cmp{ 19 bool operator()(dott &a,dott&b){ 20 return a.v>b.v; 21 } 22 }; 23 24 dott made(int x,int y){ 25 dott mid; 26 mid.num=x; 27 mid.v=y; 28 return mid; 29 } 30 31 void dij(){ 32 priority_queue<dott,vector<dott>,cmp >q; 33 memset(dis,127,sizeof(dis)); 34 dis[1]=0; 35 q.push(made(1,0)); 36 while(!q.empty()){ 37 dott mid=q.top(); 38 q.pop(); 39 if (mid.v>dis[mid.num]) continue; 40 int minn=mid.num; 41 for (int j=0;j<E[minn].size();j++){ 42 if (W[minn][j]+dis[minn]<dis[E[minn][j]]){ 43 dis[E[minn][j]]=W[minn][j]+dis[minn]; 44 q.push(made(E[minn][j],dis[E[minn][j]])); 45 } 46 } 47 } 48 } 49 50 51 int main(){ 52 scanf("%d",&n); 53 int u,v,w1; 54 for (int i=1;i<=n;i++){ 55 for (int j=1;j<=n;j++) scanf("%d",&map[i][j]); 56 } 57 58 for (int i=1;i<=n;i++) 59 for (int j=1;j<i;j++) { 60 add(j,i,map[i][j]); 61 add(i,j,map[i][j]); 62 } 63 dij(); 64 printf("%d",dis[n]); 65 }
SPFA
1 #include<cstdio> 2 #include<cstring> 3 #include<vector> 4 using namespace std; 5 const int MAXN=100; 6 vector<int> E[MAXN],V[MAXN]; 7 bool vis[MAXN]; 8 int s=1,u,v,w,n,m,dis[MAXN],b[MAXN*2]; 9 void add(int u,int v,int w){ 10 E[u].push_back(v); 11 V[u].push_back(w); 12 } 13 14 void spfa(){ 15 int h=0,t=1; 16 memset(dis,127,sizeof(dis)); 17 dis[s]=0; 18 b[t]=s; 19 while (h<t){ 20 h++; 21 int now=b[h]; 22 vis[now]=0; 23 for (int i=0;i<E[now].size();i++){ 24 int mid=E[now][i]; 25 if (dis[mid]>dis[now]+V[now][i]){ 26 dis[mid]=dis[now]+V[now][i]; 27 if (!vis[mid]){ 28 t++; 29 b[t]=mid; 30 vis[t]=1; 31 } 32 } 33 } 34 } 35 } 36 37 int main(){ 38 scanf("%d%d",&n,&m); 39 for (int i=0;i<m;i++){ 40 scanf("%d%d%d",&u,&v,&w); 41 add(u,v,w); 42 add(v,u,w); 43 } 44 spfa(); 45 printf("%d",dis[n]); 46 return 0; 47 }
