UVA 11324 The Largest Clique （强连通缩点 + DAG最长路）

链接 ： http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=30726
题意 ： 有向图G，求一个最大的点集，使得点集中任意两个节点u和v，满足 要么u可以到达v，要么v可以到达u，或者u和v可以相互到达。

可以强连通缩点成一张DAG，以为每个强连通分量要么选要么不选。求DAG上的最长路 二次建图 用了2种不同的方法，也分别用了记忆花搜索DP和直接递推DP

vector建图和记忆化搜索：

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <stack>
#include <cmath>
#include <map>
#define lson o<<1,l,m
#define rson o<<1|1,m+1,r
#define mem(a) memset(a,0,sizeof(a))
typedef long long ll;
const int N = 1005;
const int M = 50005;
const ll mod = 1000000007;

using namespace std;

int n, m, T;
int he
;

struct C {
int ne, to;
} e[M];

void add(int id, int x, int y) {
e[id].to = y;
e[id].ne = he[x];
he[x] = id;
}

int pre
, low
, scc
, dfs_clock, scc_cnt;
stack <int> S;

void dfs(int u) {

pre[u] = low[u] = ++ dfs_clock;
S.push(u);
for(int i = he[u]; i != -1; i = e[i].ne) {

int v = e[i].to;
if(pre[v] == 0) {
dfs(v);
low[u] = min(low[u], low[v]);
} else if(scc[v] == 0) {
low[u] = min(low[u], pre[v]);
}

}

if(low[u] == pre[u]) {
scc_cnt ++;
while(1) {
int x = S.top(); S.pop();
scc[x] = scc_cnt;
if(x == u) break;
}
}

}

void find_scc() {
mem(scc);
mem(pre);
dfs_clock = scc_cnt = 0;
for(int i = 1; i <= n; i++) {
if(pre[i] == 0) dfs(i);
}
}

int dp
, vis
, w
;
vector <int> G
;

int DP(int u) {
int& ans = dp[u];
if(vis[u] == 1) return ans;
vis[u] = 1;
ans = w[u];
for(int i = 0; i < G[u].size(); i++) {
int v = G[u][i];
ans = max(ans, DP(v) + w[u]);
}
return ans;
}

int main() {

cin >> T;
while(T--) {

cin >> n >> m;
memset(he, -1, sizeof(he));

for(int i = 1; i <= m; i++) {
int x, y;
scanf("%d%d", &x, &y);
add(i, x, y);
}

find_scc();
mem(w);
for(int i = 1; i <= n; i++) {
int x = scc[i];
w[x]++;
}

for(int i = 1; i <= scc_cnt; i++) {
G[i].clear();
}

int id = 1;
for(int u = 1; u <= n; u++) {
for(int i = he[u]; i != -1; i = e[i].ne) {
int v = e[i].to;
if(scc[v] != scc[u]) {
G[scc[u]].push_back(scc[v]);
}
}
}

memset(vis, 0, sizeof(vis));
int ans = 0;
for(int i = 1; i <= scc_cnt; i++) {
ans = max(ans, DP(i));
}

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

}

return 0;
}

下面是 数组邻接表建图 递推DP（类似最长上升子序列）

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <stack>
#include <cmath>
#include <map>
#define lson o<<1,l,m
#define rson o<<1|1,m+1,r
#define mem(a) memset(a,0,sizeof(a))
typedef long long ll;
const int N = 1005;
const int M = 50005;
const ll mod = 1000000007;

using namespace std;

int n, m, T;
int he
;

struct C {
int ne, to;
} e[M];

void add(int id, int x, int y) {
e[id].to = y;
e[id].ne = he[x];
he[x] = id;
}

int pre
, low
, scc
, dfs_clock, scc_cnt;
stack <int> S;

void dfs(int u) {

pre[u] = low[u] = ++ dfs_clock;
S.push(u);
for(int i = he[u]; i != -1; i = e[i].ne) {

int v = e[i].to;
if(pre[v] == 0) {
dfs(v);
low[u] = min(low[u], low[v]);
} else if(scc[v] == 0) {
low[u] = min(low[u], pre[v]);
}

}

if(low[u] == pre[u]) {
scc_cnt ++;
while(1) {
int x = S.top(); S.pop();
scc[x] = scc_cnt;
if(x == u) break;
}
}

}

void find_scc() {
mem(scc);
mem(pre);
dfs_clock = scc_cnt = 0;
for(int i = 1; i <= n; i++) {
if(pre[i] == 0) dfs(i);
}
}

int w
, he1
;
struct C1 {
int ne, to;
} e1
;
void add1(int id, int x, int y) {
e1[id].to = y;
e1[id].ne = he1[x];
he1[x] = id;
}

int dp
, vis
;

int DP(int u) {
int& ans = dp[u];
if(vis[u] == 1) return ans;
vis[u] = 1;
ans = w[u];
for(int i = he1[u]; i != -1; i = e1[i].ne) {
int v = e1[i].to;
ans = max(ans, DP(v) + w[v]);
}
return ans;
}

int main() {

cin >> T;
while(T--) {

cin >> n >> m;
memset(he, -1, sizeof(he));

for(int i = 1; i <= m; i++) {
int x, y;
scanf("%d%d", &x, &y);
add(i, x, y);
}

find_scc();
mem(w);
for(int i = 1; i <= n; i++) {
int x = scc[i];
w[x]++;
}

map < pair<int, int>, int > mp;
int id = 1;
memset(he1, -1, sizeof(he1));
for(int u = 1; u <= n; u++) {
for(int i = he[u]; i != -1; i = e[i].ne) {
int v = e[i].to;
if(scc[v] != scc[u]) {
if(mp[ make_pair(scc[u], scc[v]) ] == 0) {
mp[ make_pair(scc[u], scc[v]) ] = 1;
add1(id, scc[u], scc[v]);
id++;
}
}
}
}

memset(vis, 0, sizeof(vis));
int ans = 0;

for(int i = 1; i <= scc_cnt; i++) {
dp[i] = w[i];
for(int j = 1; j < i; j++) {
int u;
for(u = he1[i]; u != -1; u = e1[u].ne) {
if(e1[u].to == j) {
break;
}
}
if(u == -1) continue;
dp[i] = max(dp[i], dp[j] + w[i]);
}
ans = max(ans, dp[i]);
}

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

}

return 0;
}