HDU2874 Connections between cities LCA+并查集
给出一个森林,问u到v的最短距离
可以先对整个图做一次并查集求出个联通分块
对于每一个联通块进行一次Tarjan,不在一个块里面的,ans = -1;
这样写就是复杂度还高一点。。
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 using namespace std; 5 const int maxn = 20005; 6 struct edge 7 { 8 int v,w,next; 9 }e[maxn]; 10 struct node 11 { 12 int v,ans,next; 13 }q[maxn*100]; 14 int cnt,cnt_a,h1[maxn],h2[maxn],d[maxn]; 15 int fa[maxn],fat[maxn],vis[maxn],vis1[maxn]; 16 void init(int n) 17 { 18 memset(h1,-1,sizeof(h1)); 19 memset(h2,-1,sizeof(h2)); 20 memset(vis,0,sizeof(vis)); 21 memset(vis1,0,sizeof(vis1)); 22 cnt = 0;cnt_a = 0; 23 for(int i = 1;i<=n;++i)fa[i] = i,fat[i] = i; 24 } 25 int findd(int x) 26 { 27 int ret = x; 28 while(ret!=fa[ret])ret = fa[ret]; 29 while(x!=ret) 30 { 31 int t = fa[x]; 32 fa[x] = ret; 33 x = t; 34 } 35 return ret; 36 } 37 int finddd(int x) 38 { 39 int ret = x; 40 while(ret!=fat[ret])ret=fat[ret]; 41 while(x!=ret) 42 { 43 int t = fat[x]; 44 fat[x] = ret; 45 x = t; 46 } 47 return ret; 48 } 49 void add(int u,int v,int w) 50 { 51 e[cnt].v = v;e[cnt].w = w; 52 e[cnt].next = h1[u]; 53 h1[u] = cnt++; 54 } 55 void add_a(int u,int v) 56 { 57 q[cnt_a].v = v; 58 q[cnt_a].next = h2[u]; 59 h2[u] = cnt_a++; 60 } 61 void lca(int rt) 62 { 63 vis[rt] = 1; 64 for(int i = h1[rt];i!=-1;i=e[i].next)if(!vis[e[i].v]) 65 { 66 d[e[i].v] = d[rt]+e[i].w; 67 lca(e[i].v); 68 fat[e[i].v] = rt; 69 } 70 vis1[rt] = 1; 71 for(int i = h2[rt];i!=-1;i=q[i].next) 72 if(findd(q[i].v)!=findd(rt))q[i].ans = -1; 73 else 74 { 75 if(vis1[q[i].v]) 76 q[i].ans = q[i^1].ans = d[rt]+d[q[i].v]-2*d[finddd(q[i].v)]; 77 } 78 } 79 int main() 80 { 81 // freopen("in.txt","r",stdin); 82 int n,m,k; 83 while(~scanf("%d%d%d",&n,&m,&k)) 84 { 85 init(n); 86 for(int i = 1;i<=m;++i) 87 { 88 int x,y,z;scanf("%d%d%d",&x,&y,&z); 89 add(x,y,z); 90 add(y,x,z); 91 fa[findd(x)] = findd(y); 92 } 93 94 for(int i = 1;i<=k;++i) 95 { 96 int u,v;scanf("%d%d",&u,&v); 97 add_a(u,v); 98 add_a(v,u); 99 } 100 for(int i = 1;i<=n;++i)if(fa[i]==i) 101 lca(i); 102 for(int i = 0;i<cnt_a;i+=2)//printf("i=%d %d\n",i,q[i].ans); 103 if(q[i].ans==-1)printf("Not connected\n"); 104 else printf("%d\n",q[i].ans); 105 } 106 return 0; 107 }
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