标签:
给n个城市, m条边, q个询问, 每个询问, 输出城市a和b的最短距离, 如果不联通, 输出not connected。
用并查集判联通, 如果不连通, 那么两个联通块之间加一条权值很大的边。 然后树链剖分.....
#include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <map> #include <set> #include <string> #include <queue> #include <stack> #include <bitset> using namespace std; #define pb(x) push_back(x) #define ll long long #define mk(x, y) make_pair(x, y) #define lson l, m, rt<<1 #define mem(a) memset(a, 0, sizeof(a)) #define rson m+1, r, rt<<1|1 #define mem1(a) memset(a, -1, sizeof(a)) #define mem2(a) memset(a, 0x3f, sizeof(a)) #define rep(i, n, a) for(int i = a; i<n; i++) #define fi first #define se second typedef pair<int, int> pll; const double PI = acos(-1.0); const double eps = 1e-8; const int mod = 1e9+7; const int inf = 1061109567; const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; const int maxn = 10005; int head[maxn*2], fa[maxn], son[maxn], sz[maxn], deep[maxn], top[maxn], w[maxn], f[maxn], cnt, num; ll sum[maxn<<2]; struct node { int to, nextt, w; }e[maxn*2]; struct ed { int u, v; ll w; ed(){} ed(int u, int v, ll w):u(u),v(v),w(w){} }edge[maxn]; void init() { mem1(head); num = cnt = 0; } void add(int u, int v, int w) { e[num].to = v, e[num].nextt = head[u], e[num].w = w, head[u] = num++; } void dfs1(int u, int fa) { sz[u] = 1; deep[u] = deep[fa]+1; son[u] = -1; f[u] = fa; for(int i = head[u]; ~i; i = e[i].nextt) { int v = e[i].to; if(v == fa) continue; dfs1(v, u); sz[u] += sz[v]; if(son[u]==-1||sz[v]>sz[son[u]]) son[u] = v; } } void dfs2(int u, int tp) { w[u] = ++cnt, top[u] = tp; if(~son[u]) dfs2(son[u], tp); for(int i = head[u]; ~i; i = e[i].nextt) { int v = e[i].to; if(v == f[u]||v == son[u]) continue; dfs2(v, v); } } void pushUp(int rt) { sum[rt] = sum[rt<<1]+sum[rt<<1|1]; } void update(int p, ll val, int l, int r, int rt) { if(l == r) { sum[rt] = val; return ; } int m = l+r>>1; if(p<=m) update(p, val, lson); else update(p, val, rson); pushUp(rt); } ll query(int L, int R, int l, int r, int rt) { if(L<=l&&R>=r) { return sum[rt]; } int m = l+r>>1; ll ret = 0; if(L<=m) ret += query(L, R, lson); if(R>m) ret += query(L, R, rson); return ret; } ll find(int u, int v) { int f1 = top[u], f2 = top[v]; ll ret = 0; while(f1 != f2) { if(deep[f1]<deep[f2]) { swap(f1, f2); swap(u, v); } ret += query(w[f1], w[u], 1, cnt, 1); u = f[f1]; f1 = top[u]; } if(u == v) return ret; if(deep[u]>deep[v]) swap(u, v); ret += query(w[son[u]], w[v], 1, cnt, 1); return ret; } int findd(int u) { return fa[u] == u?u:findd(fa[u]); } int main() { int n, m, q, x, y, z; while(cin>>n>>m>>q) { init(); int ecnt = 0; for(int i = 1; i<=n; i++) fa[i] = i; for(int i = 0; i<m; i++) { scanf("%d%d%d", &x, &y, &z); add(x, y, z); add(y, x, z); edge[ecnt++] = ed(x, y, z); x = findd(x); y = findd(y); if(x!=y) fa[x] = y; } x = findd(1); for(int i = 2; i<=n; i++) { y = findd(i); if(y!=x) { fa[y] = x; edge[ecnt++] = ed(x, y, (ll)1e12); add(x, y, inf); add(y, x, inf); } } dfs1(1, 0); dfs2(1, 1); for(int i = 0; i<ecnt; i++) { if(deep[edge[i].u]>deep[edge[i].v]) { swap(edge[i].u, edge[i].v); } update(w[edge[i].v], edge[i].w, 1, cnt, 1); } while(q--) { scanf("%d%d", &x, &y); ll ans = find(x, y); if(ans>=1e12) { puts("Not connected"); } else { printf("%d\n", ans); } } } return 0; }
hdu 2874Connections between cities LCA
标签:
原文地址:http://www.cnblogs.com/yohaha/p/5094600.html