标签：== point center sizeof n+1 nod node tail include
ccf20170304地铁修建_Solution
这里最短路为所以从点1到点n的路径中最长的道路的长度。
因为1 ≤ n ≤ 100000，1 ≤ m ≤ 200000，属于稀疏图，所以使用Spfa(循环队列)较适合，如果使用dijkstra需要堆优化。
其实这道题用并查集最好，对所有道路长度从小到大排序，依次添加道路，直到点1和点n相连(属于同一个集合)。
1.并查集
1 #include <stdio.h> 2 #include <stdlib.h> 3 #define maxn 100000 4 #define maxm 200000 5 6 //并查集 UnionFind 7 //171s 8 9 struct node 10 { 11 long x,y,len; 12 }; 13 14 long father[maxn+1]; 15 16 long cmp(const void *a,const void *b) 17 { 18 return (*(struct node *)a).len-(*(struct node *)b).len; 19 } 20 21 long getfather(long d) 22 { 23 if (father[d]==d) 24 return d; 25 father[d]=getfather(father[d]); 26 return father[d]; 27 } 28 29 int main() 30 { 31 long n,m,i,u,v; 32 struct node *road=(struct node *) malloc (sizeof(struct node)*(maxm+1)); 33 scanf("%ld%ld",&n,&m); 34 for (i=1;i<=m;i++) 35 scanf("%ld%ld%ld",&road[i].x,&road[i].y,&road[i].len); 36 qsort(&road[1],m,sizeof(struct node),cmp); 37 for (i=1;i<=n;i++) 38 father[i]=i; 39 //边从小到大 加入图中，直到1到n可达 40 //每次加边(x,y) father[y]=x 41 for (i=1;i<=m;i++) 42 { 43 u=getfather(road[i].x); 44 v=getfather(road[i].y); 45 father[v]=u; 46 47 //每一次判断1,n的父亲是否相同，如果是，即1到n可达 48 if (getfather(father[1])==getfather(father[n])) 49 break; 50 } 51 //即1到n必经过该边i，而该边长度最长 52 printf("%ld\n",road[i].len); 53 return 0; 54 }
2.Spfa(循环队列)
1 #include <stdio.h> 2 #include <stdlib.h> 3 #include <malloc.h> 4 #include <stdbool.h> 5 #define maxn 100000 6 #define maxm 200000 7 #define maxc 1000000 8 9 long max(long a,long b) 10 { 11 if (a>b) 12 return a; 13 else 14 return b; 15 } 16 17 int main() 18 { 19 struct node 20 { 21 long d,len; 22 struct node *next; 23 }; 24 long n,m,head,tail,d,value,i,a,b,c; 25 long *dis=(long *) malloc (sizeof(long)*(maxn+1)); 26 //循环队列 27 long *line=(long *) malloc (sizeof(long)*(maxn+1)); 28 bool *vis=(bool *) malloc (sizeof(bool)*(maxn+1)); 29 //本身已经是struct node *point，创建数组再加"*" 30 struct node **point=(struct node **) malloc (sizeof(struct node *)*(maxn+1)); 31 struct node *t; 32 scanf("%ld%ld",&n,&m); 33 34 for (i=1;i<=n;i++) 35 { 36 point[i]=NULL; 37 //1 ≤ c ≤ 1000000 38 //max<=c 39 dis[i]=maxc; 40 vis[i]=true; 41 } 42 for (i=1;i<=m;i++) 43 { 44 scanf("%ld%ld%ld",&a,&b,&c); 45 //build a 46 t=(struct node *) malloc (sizeof(struct node)); 47 t->d=b; 48 t->len=c; 49 if (point[a]!=NULL) 50 t->next=point[a]; 51 else 52 t->next=NULL; 53 point[a]=t; 54 //build b 55 t=(struct node *) malloc (sizeof(struct node)); 56 t->d=a; 57 t->len=c; 58 if (point[b]!=NULL) 59 t->next=point[b]; 60 else 61 t->next=NULL; 62 point[b]=t; 63 } 64 dis[1]=0; 65 vis[1]=false; 66 line[1]=1; 67 //head=front-1 牺牲一个位置 front为队列头位置 68 head=0; 69 tail=1; 70 //这里的循环队列不用判断空或者溢出 71 //因为如果那样的话，已经不能用了。 72 //不存在空的情况。而数组要开的足够大，使队列不溢出。 73 while (head!=tail) 74 { 75 //head++; 76 //队列为0~n 77 head++; 78 if (head==n+1) 79 head=0; 80 d=line[head]; 81 t=point[d]; 82 while (t) 83 { 84 if (dis[d]<t->len) 85 value=t->len; 86 else 87 value=dis[d]; 88 if (value<dis[t->d]) 89 { 90 dis[t->d]=value; 91 //如果该点未被放在队列里，则入队；否则不入队 92 //即在队列里的点都不重复 93 //则队列(tail-head)最多有n+1个数(n个点+空位置(head)) 94 if (vis[t->d]) 95 { 96 //tail++; 97 tail++; 98 if (tail==n+1) 99 tail=0; 100 line[tail]=t->d; 101 vis[t->d]=false; 102 } 103 } 104 t=t->next; 105 } 106 vis[d]=true; 107 } 108 printf("%ld\n",dis[n]); 109 return 0; 110 }
3.dijkstra(80 Points)
1 #include <stdio.h> 2 #include <stdlib.h> 3 #include <malloc.h> 4 #include <stdbool.h> 5 #define maxn 100000 6 #define maxm 200000 7 #define maxc 1000000 8 #define maxdist 1000000000 9 10 //裸dijkstra 80分 超时 11 12 long max(long a,long b) 13 { 14 if (a>b) 15 return a; 16 else 17 return b; 18 } 19 20 int main() 21 { 22 struct node 23 { 24 long d,len; 25 struct node *next; 26 }; 27 struct node **point=(struct node **) malloc (sizeof(struct node *)*(maxn+1)); 28 struct node *p; 29 bool *vis=(bool *) malloc (sizeof(bool)*(maxn+1)); 30 long *dist=(long *) malloc (sizeof(long)*(maxn+1)); 31 long n,m,i,j,a,b,c,d,value,mindist; 32 scanf("%ld%ld",&n,&m); 33 for (i=1;i<=n;i++) 34 { 35 point[i]=NULL; 36 dist[i]=maxc; 37 vis[i]=true; 38 } 39 for (i=1;i<=m;i++) 40 { 41 scanf("%ld%ld%ld",&a,&b,&c); 42 //build a 43 p=(struct node *) malloc (sizeof(struct node)); 44 p->d=b; 45 p->len=c; 46 if (point[a]!=NULL) 47 p->next=point[a]; 48 else 49 p->next=NULL; 50 point[a]=p; 51 //build b 52 p=(struct node *) malloc (sizeof(struct node)); 53 p->d=a; 54 p->len=c; 55 if (point[b]!=NULL) 56 p->next=point[b]; 57 else 58 p->next=NULL; 59 point[b]=p; 60 } 61 for (i=1;i<=n;i++) 62 { 63 vis[i]=true; 64 dist[i]=maxdist; 65 } 66 //从点1开始 67 dist[1]=0; 68 d=1; 69 vis[1]=false; 70 for (i=1;i<n;i++) 71 { 72 p=point[d]; 73 while (p) 74 { 75 if (dist[d]<p->len) 76 value=p->len; 77 else 78 value=dist[d]; 79 if (value<dist[p->d]) 80 dist[p->d]=value; 81 p=p->next; 82 } 83 mindist=maxdist; 84 for (j=2;j<=n;j++) 85 if (vis[j] && dist[j]<mindist) 86 { 87 mindist=dist[j]; 88 d=j; 89 } 90 //从点n结束 91 if (d==n) 92 break; 93 vis[d]=false; 94 } 95 printf("%ld\n",dist[n]); 96 //最小堆优化 97 return 0; 98 }
4.dijkstra堆优化
1 #include <stdio.h> 2 #include <stdlib.h> 3 #include <malloc.h> 4 #include <stdbool.h> 5 #define maxn 100000 6 #define maxm 200000 7 #define maxc 1000000 8 #define maxdist 1000000000 9 10 //裸dijkstra 80分 超时 11 12 long max(long a,long b) 13 { 14 if (a>b) 15 return a; 16 else 17 return b; 18 } 19 20 int main() 21 { 22 struct node 23 { 24 long d,len; 25 struct node *next; 26 }; 27 struct node **point=(struct node **) malloc (sizeof(struct node *)*(maxn+1)); 28 struct node *p; 29 bool *vis=(bool *) malloc (sizeof(bool)*(maxn+1)); 30 long *dist=(long *) malloc (sizeof(long)*(maxn+1)); 31 long n,m,i,j,a,b,c,d,value,mindist; 32 scanf("%ld%ld",&n,&m); 33 for (i=1;i<=n;i++) 34 { 35 point[i]=NULL; 36 dist[i]=maxc; 37 vis[i]=true; 38 } 39 for (i=1;i<=m;i++) 40 { 41 scanf("%ld%ld%ld",&a,&b,&c); 42 //build a 43 p=(struct node *) malloc (sizeof(struct node)); 44 p->d=b; 45 p->len=c; 46 if (point[a]!=NULL) 47 p->next=point[a]; 48 else 49 p->next=NULL; 50 point[a]=p; 51 //build b 52 p=(struct node *) malloc (sizeof(struct node)); 53 p->d=a; 54 p->len=c; 55 if (point[b]!=NULL) 56 p->next=point[b]; 57 else 58 p->next=NULL; 59 point[b]=p; 60 } 61 for (i=1;i<=n;i++) 62 { 63 vis[i]=true; 64 dist[i]=maxdist; 65 } 66 //从点1开始 67 dist[1]=0; 68 d=1; 69 vis[1]=false; 70 for (i=1;i<n;i++) 71 { 72 p=point[d]; 73 while (p) 74 { 75 if (dist[d]<p->len) 76 value=p->len; 77 else 78 value=dist[d]; 79 if (value<dist[p->d]) 80 dist[p->d]=value; 81 p=p->next; 82 } 83 mindist=maxdist; 84 for (j=2;j<=n;j++) 85 if (vis[j] && dist[j]<mindist) 86 { 87 mindist=dist[j]; 88 d=j; 89 } 90 //从点n结束 91 if (d==n) 92 break; 93 vis[d]=false; 94 } 95 printf("%ld\n",dist[n]); 96 //最小堆优化 97 return 0; 98 }
标签：== point center sizeof n+1 nod node tail include
原文地址：http://www.cnblogs.com/cmyg/p/6816470.html
