UVA 11324 - The Largest Clique（SCC缩点 + DP）

题目链接：点击打开链接

思路：如果没有环， 该题就是DAG上最长路， 现在有环， 我们把强连通分量缩点， 那么缩点之后的图就是一个DAG， 可以用DP求解， 用d[i]表示以i结尾的最长路距离。

细节参见代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
typedef long double ld;
const ld eps = 1e-9, PI = 3.1415926535897932384626433832795;
const int mod = 1000000000 + 7;
const int INF = 0x3f3f3f3f;
// & 0x7FFFFFFF
const int seed = 131;
const ll INF64 = ll(1e18);
const int maxn = 1e3 + 10;
int T,n,m,u,v,pre[maxn],kase=0,lowlink[maxn], sccno[maxn], dfs_clock, scc_cnt,d[maxn], vis[maxn],cnt[maxn];
vector<int> g[maxn], G[maxn ];
stack<int> S;
void dfs(int u) {
    pre[u] = lowlink[u] = ++dfs_clock;
    S.push(u);
    int len = g[u].size();
    for(int i = 0; i < len; i++) {
        int v = g[u][i];
        if(!pre[v]) {
            dfs(v);
            lowlink[u] = min(lowlink[u], lowlink[v]);
        }
        else if(!sccno[v]) {
            lowlink[u] = min(lowlink[u], pre[v]);
        }
    }
    if(lowlink[u] == pre[u]) {
        scc_cnt++;
        for(;;) {
            int x = S.top(); S.pop();
            sccno[x] = scc_cnt;
            if(x == u) break;
        }
    }
}
void find_scc(int n) {
    for(int i = 1; i <= n; i++) {
        if(!pre[i]) dfs(i);
    }
}
void init(int n) {
    dfs_clock = scc_cnt = 0;
    for(int i = 1; i <= n; i++) {
        g[i].clear();
        sccno[i] = pre[i] = 0;
    }
}
int dp(int i) {
    int& ans = d[i];
    if(vis[i] == kase) return ans;
    vis[i] = kase;
    int len = G[i].size();
    if(len == 0) return ans = cnt[i];
    ans = 0;
    for(int j = 0; j < len; j++) {
        int v = G[i][j];
        ans = max(ans, dp(v) + cnt[i]);
    }
    return ans;
}
int main() {
    scanf("%d",&T);
    while(T--) {
        scanf("%d%d",&n,&m);
        init(n);
        for(int i = 1; i <= m; i++) {
            scanf("%d%d",&u,&v);
            g[u].push_back(v);
        }
        find_scc(n);
        for(int i = 1; i <= scc_cnt; i++) G[i].clear(), cnt[i] = 0;
        for(int i = 1; i <= n; i++) cnt[sccno[i]]++;

        for(int u = 1; u <= n; u++) {
            int len = g[u].size();
            for(int j = 0; j < len; j++) {
                int v = g[u][j];
                if(sccno[u] != sccno[v]) G[sccno[u]].push_back(sccno[v]);
            }
        }
        int ans = 0;    ++kase;
        for(int i = 1; i <= scc_cnt; i++) {
            int cur = dp(i);
            ans = max(ans, cur);
        }
        printf("%d\n", ans);
    }
    return 0;
}