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Description
Input
Output
Sample Input
7 6 1 6 13 E 6 3 9 E 3 5 7 S 4 1 3 N 2 4 20 W 4 7 2 S 3 1 6 1 4 2 6
Sample Output
13 3 36
Hint
#include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring> #include<vector> #include<list> #include<set> #include<map> #define true ture #define false flase using namespace std; #define ll long long #define inf 0xfffffff int scan() { int res = 0 , ch ; while( !( ( ch = getchar() ) >= ‘0‘ && ch <= ‘9‘ ) ) { if( ch == EOF ) return 1 << 30 ; } res = ch - ‘0‘ ; while( ( ch = getchar() ) >= ‘0‘ && ch <= ‘9‘ ) res = res * 10 + ( ch - ‘0‘ ) ; return res ; } #define maxn 100010 #define M 22 struct is { int v,next,w; } edge[maxn*2]; int deep[maxn],jiedge; int dis[maxn]; int head[maxn]; int rudu[maxn]; int fa[maxn][M]; void add(int u,int v,int w) { jiedge++; edge[jiedge].v=v; edge[jiedge].w=w; edge[jiedge].next=head[u]; head[u]=jiedge; } void dfs(int u) { for(int i=head[u]; i; i=edge[i].next) { int v=edge[i].v; int w=edge[i].w; if(!deep[v]) { dis[v]=dis[u]+edge[i].w; deep[v]=deep[u]+1; fa[v][0]=u; dfs(v); } } } void st(int n) { for(int j=1; j<M; j++) for(int i=1; i<=n; i++) fa[i][j]=fa[fa[i][j-1]][j-1]; } int LCA(int u , int v) { if(deep[u] < deep[v]) swap(u , v) ; int d = deep[u] - deep[v] ; int i ; for(i = 0 ; i < M ; i ++) { if( (1 << i) & d ) // 注意此处,动手模拟一下,就会明白的 { u = fa[u][i] ; } } if(u == v) return u ; for(i = M - 1 ; i >= 0 ; i --) { if(fa[u][i] != fa[v][i]) { u = fa[u][i] ; v = fa[v][i] ; } } u = fa[u][0] ; return u ; } void init() { memset(head,0,sizeof(head)); memset(fa,0,sizeof(fa)); memset(rudu,0,sizeof(rudu)); memset(deep,0,sizeof(deep)); jiedge=0; } int main() { int x,n,t; while(~scanf("%d%d",&n,&x)) { init(); for(int i=0; i<x; i++) { char a[2]; int u,v,w; scanf("%d%d%d %s",&u,&v,&w,a); add(u,v,w); add(v,u,w);//双向可以从任意点开始,并且避免有环 } deep[1]=1; dis[1]=0; dfs(1); st(n); scanf("%d",&t); while(t--) { int a,b; scanf("%d%d",&a,&b); printf("%d\n",dis[a]-2*dis[LCA(a,b)]+dis[b]); } } return 0; }
poj 1986 Distance Queries 带权lca 模版题
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原文地址:http://www.cnblogs.com/jhz033/p/5406487.html