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bzoj1093 [ZJOI2007]最大半联通子图

时间:2018-09-14 23:59:04      阅读:64      评论:0      收藏:0      [点我收藏+]

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最大半联通子图对应缩点后的$DAG$上的最长链

复杂度$O(n + m)$

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

extern inline char gc() {
    static char RR[23456], *S = RR + 23333, *T = RR + 23333;
    if(S == T) fread(RR, 1, 23333, stdin), S = RR;
    return *S ++;
}
inline int read() {
    int p = 0, w = 1; char c = gc();
    while(c > 9 || c < 0) { if(c == -) w = -1; c = gc(); }
    while(c >= 0 && c <= 9) p = p * 10 + c - 0, c = gc();
    return p * w; 
}

#define ri register int
#define sid 1005000

int n, m, id, nid, cnp, mod, top;
int pre[sid], nxt[sid], node[sid], cap[sid], vis[sid];
int low[sid], dfn[sid], st[sid], ins[sid], cnt[sid], b[sid], deg[sid], q[sid];

inline void addedge(int u, int v) {
    nxt[++ cnp] = cap[u]; cap[u] = cnp; 
    pre[cnp] = u; node[cnp] = v; deg[v] ++;
}

void tarjan(int o, int fa) {
    low[o] = dfn[o] = ++ id; st[++ top] = o; ins[o] = 1;
    #define cur node[i]
    for(int i = cap[o]; i; i = nxt[i]) {
        if(!dfn[cur]) tarjan(cur, o), low[o] = min(low[o], low[cur]);
        else if(ins[cur]) low[o] = min(low[o], dfn[cur]);
    }
    if(dfn[o] == low[o]) {
        int p; ++ nid;
        do{ p = st[top --]; b[p] = nid; 
            ins[p] = 0; cnt[nid] ++; 
        } while(p != o);
    }
}

inline void inc(int &a, int b)
{ a += b; if(a >= mod) a -= mod; }

struct dp {
    int sz, num;
    friend void cmax(dp &a, dp b) {
        if(b.sz > a.sz) a = b;
        else if(b.sz == a.sz) inc(a.num, b.num);
    }
} f[sid];

void top_dp() {
    int fr = 1, to = 0;
    for(ri i = 1; i <= nid; i ++) {
        if(!deg[i]) q[++ to] = i;
        f[i] = { cnt[i], 1 };
    }
    #define cur node[i]
    while(fr <= to) {
        int o = q[fr ++];
        for(ri i = cap[o]; i; i = nxt[i]) {
            deg[cur] --; if(!deg[cur]) q[++ to] = cur;
            if(vis[cur] == o) continue;
            cmax(f[cur], (dp){ f[o].sz + cnt[cur], f[o].num } );
            vis[cur] = o;
        }
    }
    dp ans = { 0, 0 };
    for(ri i = 1; i <= nid; i ++) cmax(ans, f[i]);
    printf("%d\n%d\n", ans.sz, ans.num);
}

int main() {
    n = read(); m = read(); mod = read();
    for(ri i = 1; i <= m; i ++) {
        int u = read(), v = read();
        addedge(u, v);
    }
    for(ri i = 1; i <= n; i ++)
    if(!dfn[i]) tarjan(i, 0);
    memset(cap, 0, (n + 2) << 2);
    memset(deg, 0, (n + 2) << 2);
    int cno = cnp; cnp = 0;
    for(ri i = 1; i <= cno; i ++)
    if(b[pre[i]] != b[node[i]]) addedge(b[pre[i]], b[node[i]]);
    top_dp();
    return 0;
}

 

bzoj1093 [ZJOI2007]最大半联通子图

标签:info   include   http   clu   void   zjoi   bubuko   ati   pac   

原文地址:https://www.cnblogs.com/reverymoon/p/9649450.html

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