HDU 3072 SCC Intelligence System

给出一个带权有向图，要使整个图连通。SCC中的点之间花费为0，所以就先缩点，然后缩点后两点之间的权值为最小边的权值，把这些权值累加起来就是答案。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <stack>
using namespace std;

int n, m;

const int maxn = 50000 + 10;

vector<int> G[maxn], C[maxn];

stack<int> S;
int pre[maxn], lowlink[maxn], sccno[maxn];
int dfs_clock, scc_cnt;

void dfs(int u)
{
    pre[u] = lowlink[u] = ++dfs_clock;
    S.push(u);
    for(int i = 0; i < G[u].size(); 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()
{
    dfs_clock = scc_cnt = 0;
    memset(pre, 0, sizeof(pre));
    memset(sccno, 0, sizeof(sccno));
    for(int i = 0; i < n; i++) if(!pre[i]) dfs(i);
}

int cost[maxn];

int main()
{
    while(scanf("%d%d", &n, &m) == 2)
    {
        for(int i = 0; i < n; i++) { G[i].clear(); C[i].clear(); }
        while(m--)
        {
            int u, v, d; scanf("%d%d%d", &u, &v, &d);
            G[u].push_back(v); C[u].push_back(d);
        }
        find_scc();

        memset(cost, -1, sizeof(cost));
        for(int i = 0; i < n; i++)
            for(int j = 0; j < G[i].size(); j++)
            {
                int u = sccno[i], v = sccno[G[i][j]];
                if(u == v) continue;
                if(cost[v] == -1) cost[v] = C[i][j];
                else cost[v] = min(cost[v], C[i][j]);
            }

        int ans = 0;
        for(int i = 1; i <= scc_cnt; i++) if(cost[i] != -1) ans += cost[i];
        printf("%d
", ans);
    }

    return 0;
}