luogu P2194 HXY烧情侣 题解

Posted 2020-12-25 misakaazusa

题目链接：https://www.luogu.org/problemnew/show/P2194
第一问：缩点并且统计其强连通分量里的最小耗费。把所有强连通分量的最小耗费加起来。
第二问：统计在每个强连通分量里与最小耗费相同的点数。乘法原理统计所有强连通分量答案。

#include <stack>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 300000 + 10;
const int inf = 0x7fffffff;
const int mod = 1e9 + 7;
struct edge{
    int from, to, next;
}e[maxn<<2];
int head[maxn], cnt;
int n, m, ans1, ans2 = 1, dfn[maxn], low[maxn], tim, color[maxn], num, val[maxn], minpay[maxn], tot[maxn];
bool vis[maxn];
stack<int> s;
void add(int u, int v)
{
    e[++cnt].from = u;
    e[cnt].next = head[u];
    e[cnt].to = v;
    head[u] = cnt;
}
void tarjan(int x)
{
    dfn[x] = low[x] = ++tim;
    vis[x] = 1; s.push(x);
    for(int i = head[x]; i != -1; i = e[i].next)
    {
        int v = e[i].to;
        if(!dfn[v])
        {
            tarjan(v);
            low[x] = min(low[x], low[v]);
        }
        else if(vis[v])
        {
            low[x] = min(low[x], low[v]);
        }
    }
    if(dfn[x] == low[x])
    {
        color[x] = ++num;
        vis[x] = 0;
        minpay[num] = min(minpay[num], val[x]);
        while(s.top() != x)
        {
            color[s.top()] = num;
            vis[s.top()] = 0;
            minpay[num] = min(minpay[num], val[s.top()]);
            s.pop();
        }
        s.pop();
    }
}
int main()
{
    memset(head, -1, sizeof(head));
    scanf("%d",&n);
    for(int i = 1; i <= n; i++) minpay[i] = inf;
    for(int i = 1; i <= n; i++) scanf("%d",&val[i]);
    scanf("%d",&m);
    for(int i = 1; i <= m; i++)
    {
        int u, v;
        scanf("%d%d",&u,&v);
        add(u,v);
    }
    for(int i = 1; i <= n; i++)
        if(!dfn[i]) tarjan(i);
    //for(int i = 1; i <= n; i++) cout<<minpay[i];
    for(int i = 1; i <= num; i++) ans1 += minpay[i];
    cout<<ans1<<" ";
    for(int i = 1; i <= n; i++)
    {
        if(val[i] == minpay[color[i]])
        tot[color[i]]++;
    }
    for(int i = 1; i <= num; i++) ans2 = (ans2*tot[i])%mod;
    cout<<ans2<<endl;
    return 0;
}