CodeForces-427C Checkposts
作者:互联网
Checkposts
\(tarjan\)
如果是 \(DAG\) 图,则只用找入度为 \(0\) 的点即可
因此考虑缩点后,找所有入度为 \(0\) 的点
最小值则为,缩点后所有入度为 \(0\) 的强连通块中,每个都拿一个代价最小的点
方案数为,在上述的强连通块,记录一下代价最小的点有多少个,全部相乘即可
因此 \(tarjan\) 缩点要维护强连通块中,代价最小的点的数量
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <stack>
#include <vector>
#include <queue>
using namespace std;
typedef long long ll;
const int maxn = 3e5 + 10;
const ll inf = 1e18;
const ll mod = 1e9 + 7;
ll val[maxn];
stack<int>st;
int vis[maxn], scc[maxn], cnt_scc = 0;
int low[maxn], dfn[maxn], tp = 0;
vector<int>gra[maxn];
ll minn[maxn], cnt[maxn];
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++;
minn[cnt_scc] = inf;
while(st.top() != now)
{
int x = st.top();
st.pop();
vis[x] = 0;
scc[x] = cnt_scc;
if(val[x] == minn[cnt_scc]) cnt[cnt_scc]++;
else if(val[x] < minn[cnt_scc]) {cnt[cnt_scc] = 1; minn[cnt_scc] = val[x];}
}
st.pop();
vis[now] = 0;
scc[now] = cnt_scc;
if(val[now] == minn[cnt_scc]) cnt[cnt_scc]++;
else if(val[now] < minn[cnt_scc]) {cnt[cnt_scc] = 1; minn[cnt_scc] = val[now];}
}
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n;
cin >> n;
for(int i=1; i<=n; i++) cin >> val[i];
int m;
cin >> m;
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);
ll a = 0, b = 1;
for(int i=1; i<=cnt_scc; i++)
{
a += minn[i];
b = cnt[i] * b % mod;
}
cout << a << " " << b << endl;
return 0;
}
标签:cnt,int,Checkposts,CodeForces,scc,427C,maxn,low,now 来源: https://www.cnblogs.com/dgsvygd/p/16579789.html