HDU-3072-IntelligenceSystem（tarjan，贪心）

链接：https://vjudge.net/problem/HDU-3072

题意：

给你n个点，1个点到另一个点连接花费c，但是如果几个点可以相互可达，则这几个点连通花费为0.

求将整个图连通的最小花费。

思路：

tarjan，求出强连通子图。

对每个子图的进点的最小值更新，再累加即可，（不过不知道为什么）

代码：

#include <iostream>
#include <memory.h>
#include <string>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <algorithm>
#include <map>
#include <queue>
#include <math.h>
#include <cstdio>
#include <set>
#include <iterator>
#include <cstring>
using namespace std;

typedef long long LL;
const int MAXN = 5e4+10;
const int INF = 0x3f3f3f3f;

struct Node
{
    int from_, to_, value_;
    Node(int from, int to, int value):from_(from), to_(to), value_(value){}
    bool operator < (const Node &that) const
    {
        return this->value_ > that.value_;
    }
};

vector<Node> G[MAXN];
stack<int> St;
int Dfn[MAXN], Low[MAXN];
int Vis[MAXN], Dis[MAXN];
int Fa[MAXN], Val[MAXN];
int Num[MAXN];
int n, m;
int times, cnt;

void Init()
{
    for (int i = 1;i <= n;i++)
        G[i].clear(), Fa[i] = i;
    memset(Dfn, 0, sizeof(Dfn));
    memset(Low, 0, sizeof(Low));
    memset(Vis, 0, sizeof(Vis));
    memset(Dis, 0, sizeof(Dis));
    memset(Num, 0, sizeof(Num));
    memset(Val, 0, sizeof(Val));
    times = cnt = 0;
}

void Tarjan(int x)
{
    Dfn[x] = Low[x] = ++times;
    Vis[x] = 1;
    St.push(x);
    for (int i = 0;i < G[x].size();i++)
    {
        int node = G[x][i].to_;
        if (Dfn[node] == 0)
        {
            Tarjan(node);
            Low[x] = min(Low[x], Low[node]);
        }
        else if (Vis[node] == 1)
            Low[x] = min(Low[x], Dfn[node]);
    }
    if (Low[x] == Dfn[x])
    {
        cnt++;
        Num[cnt] = 0;
        while (St.top() != x)
        {
            Num[cnt]++;
            Fa[St.top()] = cnt;
            Vis[St.top()] = 0;
            St.pop();
        }
        Num[cnt]++;
        Fa[St.top()] = cnt;
        Vis[St.top()] = 0;
        St.pop();
    }
}

int main()
{
    int t, cn = 0;
    while (~scanf("%d%d", &n, &m))
    {
        Init();
        int l, r, v;
        for (int i = 1;i <= m;i++)
        {
            scanf("%d%d%d", &l, &r, &v);
            l++, r++;
            G[l].emplace_back(l, r, v);
        }
        for (int i = 1;i <= n;++i)
            if (!Dfn[i])
                Tarjan(i);
        for (int i = 1;i <= cnt;i++)
            Val[i] = INF;
        for (int i = 1;i <= n;i++)
        {
            for (int j = 0;j < G[i].size();j++)
            {
                int node = G[i][j].to_;
                if (Fa[i] != Fa[node])
                    Val[Fa[node]] = min(Val[Fa[node]], G[i][j].value_);
            }
        }
        int res = 0;
        for (int i = 1;i <= cnt;i++)
            if (Val[i] != INF)
                res += Val[i];
        cout << res << endl;
    }

    return 0;
}

原文地址：https://www.cnblogs.com/YDDDD/p/10822851.html