CSES-1686 Coin Collector

Coin Collector#

典中典题了属于是

$tarjan$ 缩点 + 记忆化搜索 或 $dp$

如果是 $DAG$ 图，直接记忆化搜索或者 $dp$ 跑一边

考虑图中有强连通分量，先缩点后记忆化搜索

每个点的硬币最大值 $dp$ 为，其他能到该点的 $dp$ 最大值转移过来

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <stack>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
vector<int>gra[maxn], gra_c[maxn];
int low[maxn], dfn[maxn], tp = 0;
int scc[maxn], cnt_scc = 0;
int vis[maxn];
ll val[maxn], val_c[maxn], dp[maxn];
stack<int>st;

void tarjan(int now)
{
    vis[now] = 1;
    st.push(now);
    dfn[now] = low[now] = ++tp;
    for(int nex : gra[now])
    {
        if(dfn[nex] == 0)
        {
            tarjan(nex);
            low[now] = min(low[now], low[nex]);
        }
        else if(vis[nex])
            low[now] = min(low[now], low[nex]);
    }
    if(low[now] == dfn[now])
    {
        cnt_scc++;
        while(st.top() != now)
        {
            int x = st.top();
            st.pop();
            vis[x] = 0;
            scc[x] = cnt_scc;
            val_c[cnt_scc] += val[x];
        }
        st.pop();
        vis[now] = 0;
        scc[now] = cnt_scc;
        val_c[cnt_scc] += val[now];
    }
}

void build(int n)
{
    for(int i=1; i<=n; i++)
    {
        for(int nex : gra[i])
        {
            if(scc[i] != scc[nex])
                gra_c[scc[nex]].push_back(scc[i]);
        }
    }
}

ll dps(int now)
{
    if(dp[now]) return dp[now];
    ll ans = 0;
    for(int nex : gra_c[now])
        ans = max(ans, dps(nex));
    return dp[now] = ans + val_c[now];
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n, m;
    cin >> n >> m;
    for(int i=1; i<=n; i++) cin >> val[i];
    for(int i=0; i<m; i++)
    {
        int a, b;
        cin >> a >> b;
        gra[a].push_back(b);
    }
    for(int i=1; i<=n; i++)
        if(dfn[i] == 0) tarjan(i);
    build(n);
    ll ans = 0;
    for(int i=1; i<=cnt_scc; i++)
        ans = max(ans, dps(i));
    cout << ans << endl;
    return 0;
}