BZOJ 1093: [ZJOI2007]最大半连通子图( tarjan + dp )

WA了好多次...

先tarjan缩点, 然后题意就是求DAG上的一条最长链. dp(u) = max{dp(v)} + tot_u, edge(u,v)存在. tot_u是scc(u)的结点数. 其实就是记忆化搜一下...重边就用set判一下

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#include<cstdio>

#include<cstring>

#include<algorithm>

#include<stack>

#include<iostream>

#include<cctype>

#include<set>

 

using namespace std;

 

const int maxn = 100009;

const int INF = 0x3F3F3F3F;

 

inline int read() {

char c = getchar();

int ret = 0;

for(; !isdigit(c); c = getchar());

for(; isdigit(c); c = getchar()) ret = ret * 10 + c - '0';

return ret;

}

 

struct edge {

int to;

edge* next;

} E[1000009], *pt = E, *head[maxn];

 

void addedge(int u, int v) {

pt->to = v; pt->next = head[u]; head[u] = pt++;

}

 

int N, scc[maxn], dfn[maxn], low[maxn], tot[maxn], CK = 0, n = -1, MOD;

stack<int> S;

set<pair<int, int> > bst;

 

void tarjan(int x) {

dfn[x] = low[x] = ++CK;

S.push(x);

for(edge* e = head[x]; e; e = e->next)

if(!dfn[e->to])

tarjan(e->to), low[x] = min(low[e->to], low[x]);

else if(!~scc[e->to])

low[x] = min(low[x], dfn[e->to]);

if(dfn[x] == low[x]) {

int t; n++;

do {

t = S.top(); S.pop();

tot[n]++;

scc[t] = n;

} while(t != x);

}

}

 

namespace DP {

edge E[1000009], *pt = E, *head[maxn];

int dp[maxn], cnt[maxn];

bool vis[maxn];

void addedge(int u, int v) {

pt->to = v; pt->next = head[u]; head[u] = pt++;

}

void init() {

memset(dp, 0, sizeof dp);

memset(cnt, 0, sizeof cnt);

memset(vis, 0, sizeof vis);

}

void Dp(int x) {

if(vis[x]) return;

vis[x] = true;

for(edge* e = head[x]; e; e = e->next) {

Dp(e->to);

if(dp[e->to] > dp[x])

dp[x] = dp[e->to], cnt[x] = cnt[e->to];

else if(dp[e->to] == dp[x] && (cnt[x] += cnt[e->to]) >= MOD) 

cnt[x] -= MOD;

}

dp[x] += tot[x];

if(!cnt[x]) cnt[x] = 1;

}

void work() {

for(int i = 0; i <= n; i++) 

if(!vis[i]) Dp(i);

int ans = 0, ans0 = 0;

for(int i = 0; i <= n; i++)

if(dp[i] > ans) ans = dp[i], ans0 = cnt[i];

else if(ans == dp[i] && (ans0 += cnt[i]) >= MOD) ans0 -= MOD;

printf("%d\n%d\n", ans, ans0);

}

}

 

void AddEdge() {

for(int x = 0; x < N; x++)

for(edge* e = head[x]; e; e = e->next) 

if(scc[x] != scc[e->to] && bst.find(make_pair(scc[x], scc[e->to])) == bst.end()) {

DP::addedge(scc[x], scc[e->to]);

bst.insert(make_pair(scc[x], scc[e->to]));

}

}

 

void TARJAN() {

memset(dfn, 0, sizeof dfn);

memset(low, 0, sizeof low);

memset(scc, -1, sizeof scc);

memset(tot, 0, sizeof tot);

for(int i = 0; i < N; i++) if(!dfn[i]) tarjan(i);

}

 

void init() {

N = read();

int m = read();

MOD = read();

while(m--) {

int u = read() - 1, v = read() - 1;

addedge(u, v);

}

DP::init();

}

 

int main() {

init();

TARJAN();

DP::init();

AddEdge();

DP::work();

return 0;

}

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1093: [ZJOI2007]最大半连通子图

Time Limit: 30 Sec  Memory Limit: 162 MB
Submit: 2105  Solved: 841
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Description

Input

第一行包含两个整数N，M，X。N，M分别表示图G的点数与边数，X的意义如上文所述。接下来M行，每行两个正整数a, b，表示一条有向边(a, b)。图中的每个点将编号为1,2,3…N，保证输入中同一个(a,b)不会出现两次。

Output

应包含两行，第一行包含一个整数K。第二行包含整数C Mod X.

Sample Input

6 6 20070603
1 2
2 1
1 3
2 4
5 6
6 4

Sample Output

3
3

HINT

对于100%的数据， N ≤100000, M ≤1000000；对于100%的数据， X ≤10^8。

Source