[BZOJ1924][SDOI2010]所驼门王的宝藏（Tarjan+拓扑排序）

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=1924
容易想到，如果在第i个藏宝宫室可以传送到第j个藏宝宫室，就连一条i−>j的有向边，Tarjan强连通分量缩点之后，在新图上找出一条路径，使得这条路径上的所有点对应的强连通分量大小之和最大，这个可以按照拓扑序进行递推来实现。
但是，在建图上存在一个问题，从横天门或纵寰门引出的边可能特别多。在极端情况下，105个横天门在同一行里出现，这样时空复杂度都是无法承受的。考虑这一点进行优化：由于在同一行的横天门一定属于同一个强连通分量，所以在建边时，只需要对在同一行的横天门构建出一个环即可，不需要两两进行连边。而此行内的其他宫室，只需要从这个环中的任意一点向这个宫室连边即可。对于纵寰门也是一样。
此外，在建图的实现上有一些小技巧，具体见代码。
代码：

#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;inline int read() {    int res = 0; bool bo = 0; char c;    while (((c = getchar()) < '0' || c > '9') && c != '-');    if (c == '-') bo = 1; else res = c - 48;    while ((c = getchar()) >= '0' && c <= '9')        res = (res << 3) + (res << 1) + (c - 48);    return bo ? ~res + 1 : res;}const int N = 1e5 + 5, M = 2e6 + 5, R = 1e6 + 5;int n, ecnt, nxt[M], adj[N], go[M], dfn[N], low[N], times, sum, num[N],top, stk[N], bel[N], ecnt2, nxt2[M], adj2[N], go2[M], tot, st[N], ed[N],las[R], now[R], nex[R], row[N], cnt[N], res[N], H, T, Q[N]; bool ins[N];struct cyx {int x, y, t, id;} a[N];void add_edge(int u, int v) {    nxt[++ecnt] = adj[u]; adj[u] = ecnt; go[ecnt] = v;}void add_edge2(int u, int v) {    nxt2[++ecnt2] = adj2[u]; adj2[u] = ecnt2; go2[ecnt2] = v;}bool comp1(cyx a, cyx b) {    if (a.x != b.x) return a.x < b.x;    return a.t < b.t;}bool comp2(cyx a, cyx b) {    if (a.y != b.y) return a.y < b.y;    return a.t < b.t;}bool comp3(cyx a, cyx b) {    if (a.x != b.x) return a.x < b.x;    return a.y < b.y;}void Tarjan(int u) {    dfn[u] = low[u] = ++times; ins[stk[++top] = u] = 1;    for (int e = adj[u], v; e; e = nxt[e])        if (!dfn[v = go[e]]) {            Tarjan(v);            low[u] = min(low[u], low[v]);        }        else if (ins[v]) low[u] = min(low[u], dfn[v]);    if (dfn[u] == low[u]) {        num[bel[u] = ++sum] = 1; ins[u] = 0; int v;        while (v = stk[top--], v != u) num[bel[v] = sum]++, ins[v] = 0;    }}void topo() {    int i; H = T = 0;    for (i = 1; i <= sum; i++) if (!cnt[i]) Q[++T] = i, res[i] = num[i];    while (H < T) {        int u = Q[++H];        for (int e = adj2[u], v; e; e = nxt2[e]) {            if (!(--cnt[v = go2[e]])) Q[++T] = v;            res[v] = max(res[v], res[u] + num[v]);        }    }}int main() {    int i, j; n = read(); read(); read();    for (i = 1; i <= n; i++) a[i].x = read(), a[i].y = read(),        a[i].t = read(), a[i].id = i;    sort(a + 1, a + n + 1, comp1);    for (i = 1; i <= n;) {        int fir = 0, lst = 0;        for (j = i; j <= n && a[i].x == a[j].x; j++) if (a[j].t == 1) {            if (!fir) fir = j; lst = j;            if (j < n && a[i].x == a[j + 1].x && a[j + 1].t == 1)                add_edge(a[j].id, a[j + 1].id);        }        if (lst) {            if (lst != fir) add_edge(a[lst].id, a[fir].id);            for (j = i; j <= n && a[i].x == a[j].x; j++) if (a[j].t != 1)                add_edge(a[lst].id, a[j].id);        }        i = j;    }    sort(a + 1, a + n + 1, comp2);    for (i = 1; i <= n;) {        int fir = 0, lst = 0;        for (j = i; j <= n && a[i].y == a[j].y; j++) if (a[j].t == 2) {            if (!fir) fir = j; lst = j;            if (j < n && a[i].y == a[j + 1].y && a[j + 1].t == 2)                add_edge(a[j].id, a[j + 1].id);        }        if (lst) {            if (lst != fir) add_edge(a[lst].id, a[fir].id);            for (j = i; j <= n && a[i].y == a[j].y; j++) if (a[j].t != 2)                add_edge(a[lst].id, a[j].id);        }        i = j;    }    sort(a + 1, a + n + 1, comp3);    for (i = 1; i <= n;) {        st[++tot] = i; row[tot] = a[i].x;        for (j = i; j <= n && a[i].x == a[j].x; j++);        ed[tot] = j - 1; i = j;    }    for (i = 1; i <= tot; i++) {        if (i > 1) for (j = st[i - 1]; j <= ed[i - 1]; j++)            las[a[j].y] = a[j].id;        for (j = st[i]; j <= ed[i]; j++) now[a[j].y] = a[j].id;        if (i < tot) for (j = st[i + 1]; j <= ed[i + 1]; j++)            nex[a[j].y] = a[j].id;        for (j = st[i]; j <= ed[i]; j++) {            if (a[j].t != 3) continue;            if (i > 1 && row[i - 1] + 1 == row[i]) {                if (las[a[j].y - 1]) add_edge(a[j].id, las[a[j].y - 1]);                if (las[a[j].y]) add_edge(a[j].id, las[a[j].y]);                if (las[a[j].y + 1]) add_edge(a[j].id, las[a[j].y + 1]);            }            if (now[a[j].y - 1]) add_edge(a[j].id, now[a[j].y - 1]);            if (now[a[j].y + 1]) add_edge(a[j].id, now[a[j].y + 1]);            if (i < tot && row[i + 1] - 1 == row[i]) {                if (nex[a[j].y - 1]) add_edge(a[j].id, nex[a[j].y - 1]);                if (nex[a[j].y]) add_edge(a[j].id, nex[a[j].y]);                if (nex[a[j].y + 1]) add_edge(a[j].id, nex[a[j].y + 1]);            }        }        if (i > 1) for (j = st[i - 1]; j <= ed[i - 1]; j++)            las[a[j].y] = 0;        for (j = st[i]; j <= ed[i]; j++) now[a[j].y] = 0;        if (i < tot) for (j = st[i + 1]; j <= ed[i + 1]; j++)            nex[a[j].y] = 0;    }    for (i = 1; i <= n; i++) if (!dfn[i]) Tarjan(i);    for (i = 1; i <= n; i++) for (int e = adj[i]; e; e = nxt[e])        if (bel[i] != bel[go[e]]) add_edge2(bel[i], bel[go[e]]),            cnt[bel[go[e]]]++;    int ans = 0; topo();    for (i = 1; i <= sum; i++) ans = max(ans, res[i]);    printf("%d\n", ans);    return 0;}