【uva 534】Frogger(图论--最小瓶颈路 模版题)
题意:平面上有N个石头,给出坐标。一只青蛙从1号石头跳到2号石头,使路径上的最长便最短。输出这个值。(2≤N≤200)
解法:最小瓶颈树。而由于这题N比较小便可以用2种方法:
1.最短路径中提到过的Floyd算法,利用DP思想枚举出所有情况,O(n3)。具体关于Floyd算法的一些解释得等我过几天写一篇博文。
2.用Kruskal算法求出最小生成树再求出路径上的最长边权。具体解释见我之前的一篇博文: 关于生成树的拓展 {附【转】最小瓶颈路与次小生成树}(图论--生成树)
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 #include<cmath> 5 #include<iostream> 6 using namespace std; 7 8 const int N=210; 9 const double INF=300010.0; 10 double d[N][N]; 11 struct node{int x,y;}a[N]; 12 13 double mmin(double x,double y) {return x<y?x:y;} 14 double mmax(double x,double y) {return x>y?x:y;} 15 double dist(int i,int j) {return sqrt((a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y));} 16 int main() 17 { 18 int n,T=0; 19 while (scanf("%d",&n)!=EOF && n) 20 { 21 int i,j,k; 22 for (i=1;i<=n;i++) scanf("%d%d",&a[i].x,&a[i].y); 23 for (i=1;i<=n;i++) 24 for (j=i;j<=n;j++) 25 d[i][j]=d[j][i]=dist(i,j); 26 for (k=1;k<=n;k++)//??? 27 for (i=1;i<=n;i++) 28 for (j=1;j<=n;j++)//?i?j????1~k???????????? 29 d[i][j]=mmin(d[i][j],mmax(d[i][k],d[k][j])); 30 printf("Scenario #%d\nFrog Distance = %.3lf\n\n",++T,d[1][2]); 31 } 32 }
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 #include<cmath> 5 #include<algorithm> 6 #include<iostream> 7 using namespace std; 8 9 const int N=210,M=20010; 10 const double INF=300010.0; 11 int n,m; 12 int fa[N]; 13 struct node{int x,y;}a[N]; 14 struct edge 15 { 16 int x,y; 17 double d; 18 edge() {} 19 edge(int i,int j,double k) {x=i;y=j;d=k;} 20 }e[M]; 21 22 double mmin(double x,double y) {return x<y?x:y;} 23 double mmax(double x,double y) {return x>y?x:y;} 24 bool cmp(edge x,edge y) {return x.d<y.d;} 25 double dist(int i,int j) {return sqrt((a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y));} 26 27 int ffind(int x) 28 { 29 if (fa[x]!=x) fa[x]=ffind(fa[x]); 30 return fa[x]; 31 } 32 double Kruskal(int u,int v) 33 { 34 int i; 35 for (i=1;i<=n;i++) fa[i]=i; 36 sort(e+1,e+1+m,cmp); 37 for (i=1;i<=m;i++) 38 { 39 int x=e[i].x,y=e[i].y; 40 int xx=ffind(x),yy=ffind(y); 41 if (xx!=yy) 42 { 43 fa[xx]=yy; 44 //if (x==u||y==u||x==v||y==v)//wrong 45 if (ffind(u)==ffind(v)) return e[i].d; 46 } 47 } 48 } 49 int main() 50 { 51 int T=0; 52 while (scanf("%d",&n)!=EOF && n) 53 { 54 int i,j; 55 for (i=1;i<=n;i++) scanf("%d%d",&a[i].x,&a[i].y); 56 m=0; 57 for (i=1;i<=n;i++) 58 for (j=i+1;j<=n;j++) 59 e[++m]=edge(i,j,dist(i,j)); 60 printf("Scenario #%d\nFrog Distance = %.3lf\n\n",++T,Kruskal(1,2)); 61 } 62 }
P.S.而我之前关于代码2打的超复杂:“乖乖地”跑完一次Kruskal后删边,再建立邻接表,打算Bfs一次......