• BZOJ 2125: 最短路


    2125: 最短路
    思路:构建圆方树,然后如果两个点的lca是圆点,直接算,否则跳到环上相应的位置,再求环上两个点的最短距离。
    代码1(在重链上跳):

    #pragma GCC optimize(2)
    #pragma GCC optimize(3)
    #pragma GCC optimize(4)
    #include<bits/stdc++.h>
    using namespace std;
    #define y1 y11
    #define fi first
    #define se second
    #define pi acos(-1.0)
    #define LL long long
    //#define mp make_pair
    #define pb push_back
    #define ls rt<<1, l, m
    #define rs rt<<1|1, m+1, r
    #define ULL unsigned LL
    #define pll pair<LL, LL>
    #define pli pair<LL, int>
    #define pii pair<int, int>
    #define piii pair<pii, int>
    #define pdd pair<double, double>
    #define mem(a, b) memset(a, b, sizeof(a))
    #define debug(x) cerr << #x << " = " << x << "
    ";
    #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    //head
    
    struct Circle_Square_Tree {
        const static int N = 2e4 + 10;
        vector<pair<int, LL> > g[N];
        int fa[N], dp[N], sz[N], son[N], top[N], dfn[N], to[N], cnt, n;
        LL dis[N], cir[N]; //环的大小
        bool vis[N];//记录到方点的最短距离是否经过回边
    
        inline void dfs1(int u, int o) {
            fa[u] = o;
            sz[u] = 1;
            dp[u] = dp[o] + 1;
            for (int i = 0; i < g[u].size(); ++i) {
                int v = g[u][i].fi;
                LL w = g[u][i].se;
                if(v != o) {
                    dis[v] = dis[u] + w;
                    dfs1(v, u);
                    sz[u] += sz[v];
                    if(sz[v] > sz[son[u]]) son[u] = v;
                }
            }
        }
        inline void dfs2(int u, int t) {
            top[u] = t;
            dfn[u] = ++cnt;
            to[cnt] = u;
            if(!son[u]) return ;
            dfs2(son[u], t);
            for (int i = 0; i < g[u].size(); ++i) {
                int v = g[u][i].fi;
                if(v != fa[u] && v != son[u]) dfs2(v, v);
            }
        }
        inline void init(int _n) {
            cnt = 0;
            n = _n;
            dp[0] = 0;
            dfs1(1, 0);
            dfs2(1, 1);
        }
        inline int lca(int u, int v) {
            int fu = top[u], fv = top[v];
            while(fu != fv) {
                if(dp[fu] >= dp[fv]) u = fa[fu], fu = top[u];
                else v = fa[fv], fv = top[v];
            }
            if(dp[u] <= dp[v]) return u;
            else return v;
        }
        inline int jump(int u, int lca) {
            int res;
            while(top[u] != top[lca]) res = top[u], u = fa[top[u]];
            if(u == lca) return res;
            else return to[dfn[lca]+1];
        }
        inline LL query(int u, int v) {
            int l = lca(u, v);
            if(l <= n) return dis[u]+dis[v]-2*dis[l];
            int uu = jump(u, l), vv = jump(v, l);
            LL d1 = dis[uu]-dis[l], d2 = dis[vv]-dis[l];
            if(!vis[uu]) d1 = cir[l]-d1;
            if(!vis[vv]) d2 = cir[l]-d2;
            return dis[u]-dis[uu]+dis[v]-dis[vv]+min(abs(d1-d2), cir[l]-abs(d1-d2));
        }
    }CST;
    const int N = 1e4 + 10;
    int n, m, q, u, v, w;
    vector<pii> g[N];
    int dfn[N], low[N], fa[N], cnt = 0, tot = 0;
    LL dep[N];
    int a[N];
    inline void solve(int u, int v, int d) {
        int cnt = 0;
        LL sum = d;
        for (int i = v; i != u; i = fa[i]) sum += dep[i] - dep[fa[i]], a[++cnt] = i;
        a[++cnt] = u;
        LL dis = 0;
        ++tot;
        for (int i = cnt; i >= 1; --i) {
            LL D = min(dis, sum-dis);
            if(D == dis) CST.vis[a[i]] = true;
            else CST.vis[a[i]] = false;
            CST.g[a[i]].pb({tot, D});
            CST.g[tot].pb({a[i], D});
            dis += dep[a[i-1]]-dep[a[i]];
        }
        CST.cir[tot] = sum;
    }
    inline void tarjan(int u, int o) {
        fa[u] = o;
        dfn[u] = low[u] = ++cnt;
        for (int i = 0; i < g[u].size(); ++i) {
            int v = g[u][i].fi;
            int w = g[u][i].se;
            if(v == o) continue;
            if(!dfn[v]) {
                dep[v] = dep[u] + w;
                tarjan(v, u);
                low[u] = min(low[u], low[v]);
            }
            else low[u] = min(low[u], dfn[v]);
            if(low[v] > dfn[u]) {
                CST.g[u].pb({v, w});
                CST.g[v].pb({u, w});
            }
        }
        for (int i = 0; i < g[u].size(); ++i) {
            int v = g[u][i].fi;
            int w = g[u][i].se;
            if(v == o) continue;
            if(fa[v] != u && dfn[v] > dfn[u]) {
                solve(u, v, w);
            }
        }
    }
    int main() {
        scanf("%d %d %d", &n, &m, &q);
        for (int i = 1; i <= m; ++i) scanf("%d %d %d", &u, &v, &w), g[u].pb({v, w}), g[v].pb({u, w});
        tot = n;
        tarjan(1, 0);
        CST.init(n);
        while(q--) {
            scanf("%d %d", &u, &v);
            printf("%lld
    ", CST.query(u, v));
        }
        return 0;
    }
    
    

    代码2(倍增跳):

    #pragma GCC optimize(2)
    #pragma GCC optimize(3)
    #pragma GCC optimize(4)
    #include<bits/stdc++.h>
    using namespace std;
    #define y1 y11
    #define fi first
    #define se second
    #define pi acos(-1.0)
    #define LL long long
    //#define mp make_pair
    #define pb push_back
    #define ls rt<<1, l, m
    #define rs rt<<1|1, m+1, r
    #define ULL unsigned LL
    #define pll pair<LL, LL>
    #define pli pair<LL, int>
    #define pii pair<int, int>
    #define piii pair<pii, int>
    #define pdd pair<double, double>
    #define mem(a, b) memset(a, b, sizeof(a))
    #define debug(x) cerr << #x << " = " << x << "
    ";
    #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    //head
    
    struct Circle_Square_Tree {
        const static int N = 2e4 + 10;
        vector<pair<int, LL> > G[N];
        int dp[N], anc[N][18], n;
        LL dis[N], cir[N]; //环的大小
        bool vis[N];//记录到方点的最短距离是否经过回边
    
        vector<pii> g[N];
        int dfn[N], low[N], fa[N], cnt, tot;
        int a[N];
        inline void solve(int u, int v, int d) {
            int cnt = 0;
            LL sum = d;
            for (int i = v; i != u; i = fa[i]) sum += dis[i] - dis[fa[i]], a[++cnt] = i;
            a[++cnt] = u;
            LL DIS = 0;
            ++tot;
            for (int i = cnt; i >= 1; --i) {
                LL D = min(DIS, sum-DIS);
                if(D == DIS) vis[a[i]] = true;
                else vis[a[i]] = false;
                G[a[i]].pb({tot, D});
                G[tot].pb({a[i], D});
                DIS += dis[a[i-1]]-dis[a[i]];
            }
            cir[tot] = sum;
        }
        inline void tarjan(int u, int o) {
            fa[u] = o;
            dfn[u] = low[u] = ++cnt;
            for (int i = 0; i < g[u].size(); ++i) {
                int v = g[u][i].fi;
                int w = g[u][i].se;
                if(v == o) continue;
                if(!dfn[v]) {
                    dis[v] = dis[u] + w;
                    tarjan(v, u);
                    low[u] = min(low[u], low[v]);
                }
                else low[u] = min(low[u], dfn[v]);
                if(low[v] > dfn[u]) {
                    G[u].pb({v, w});
                    G[v].pb({u, w});
                }
            }
            for (int i = 0; i < g[u].size(); ++i) {
                int v = g[u][i].fi;
                int w = g[u][i].se;
                if(v == o) continue;
                if(fa[v] != u && dfn[v] > dfn[u]) {
                    solve(u, v, w);
                }
            }
        }
        inline void dfs(int u, int o) {
            anc[u][0] = o;
            dp[u] = dp[o] + 1;
            for (int i = 1; i < 18; ++i) anc[u][i] = anc[anc[u][i-1]][i-1];
            for (int i = 0; i < G[u].size(); ++i) {
                int v = G[u][i].fi;
                LL w = G[u][i].se;
                if(v == o) continue;
                dis[v] = dis[u] + w;
                dfs(v, u);
            }
        }
        inline void init(int _n) {
            tot = n = _n;
            cnt = 0;
            dis[0] = dp[1] = 0;
            tarjan(1, 0);
            dfs(1, 1);
        }
        inline int lca(int u, int v) {
            if(dp[u] < dp[v]) swap(u, v);
            for (int i = 17; i >= 0; --i) if(dp[anc[u][i]] >= dp[v]) u = anc[u][i];
            if(u == v) return u;
            for (int i = 17; i >= 0; --i) if(anc[u][i] != anc[v][i]) u = anc[u][i], v = anc[v][i];
            return anc[u][0];
        }
        inline int jump(int u, int lca) {
            for (int i = 17; i >= 0; --i) if(dp[anc[u][i]] > dp[lca]) u = anc[u][i];
            return u;
        }
        inline LL query(int u, int v) {
            int l = lca(u, v);
            if(l <= n) return dis[u]+dis[v]-2*dis[l];
            int uu = jump(u, l), vv = jump(v, l);
            LL d1 = dis[uu]-dis[l], d2 = dis[vv]-dis[l];
            if(!vis[uu]) d1 = cir[l]-d1;
            if(!vis[vv]) d2 = cir[l]-d2;
            return dis[u]-dis[uu]+dis[v]-dis[vv]+min(abs(d1-d2), cir[l]-abs(d1-d2));
        }
    }CST;
    int n, m, q, u, v, w;
    int main() {
        scanf("%d %d %d", &n, &m, &q);
        for (int i = 1; i <= m; ++i) scanf("%d %d %d", &u, &v, &w), CST.g[u].pb({v, w}), CST.g[v].pb({u, w});
        CST.init(n);
        while(q--) {
            scanf("%d %d", &u, &v);
            printf("%lld
    ", CST.query(u, v));
        }
        return 0;
    }
    
    
  • 相关阅读:
    面向对象的相关知识
    模块的导入
    正怎表达式在爬虫里的应用
    前端~css
    CSS知识点
    初识Html
    Python之路--协程/IO多路复用
    进程
    锁 和 线程池
    操作系统-并发-线程-进程
  • 原文地址:https://www.cnblogs.com/widsom/p/11468746.html
Copyright © 2020-2023  润新知