学步园

传送门：http://acm.hdu.edu.cn/showproblem.php?pid=1827

求联通块入度为0的数量及传递到整个图的最小花费。算是tarjan的模板题了

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#include<stack>

using namespace std;

const int N = 1005;
const int inf = 1 << 28;

struct node{
    int to, nxt;
}e[N*2];

struct pp{
    int st, ed;
}p[N*2];

int head[N], vis[N];
int scc[N], sccnum;
int cnt, index;
int pay[N];
int low[N], dfn[N];
int n, m;
stack<int> s;
int in[N];

void init()
{
    sccnum = cnt = index = 0;
    memset(head, -1, sizeof(head));
    memset(low, 0, sizeof(low));
    memset(vis, 0, sizeof(vis));
    memset(in, 0, sizeof(in));
    memset(dfn, 0, sizeof(dfn));
    while( !s.empty() )
        s.pop();
}

void add(int u, int v)
{
    e[cnt].to = v;
    e[cnt].nxt = head[u];
    head[u] = cnt++;
}

void tarjan( int u )
{
    dfn[u] = low[u] = ++index;	//memset(0的时候++index，-1就随意了)
    vis[u] = 1;
    s.push(u);
    for( int i = head[u]; ~i; i = e[i].nxt )
    {
        int to = e[i].to;
        if( !dfn[to] )
        {
            tarjan(to);
            low[u] = min(low[u], low[to]);
        }
        else if( vis[to] )
        {
            low[u] = min(low[u], dfn[to]);
        }
    }
    if( low[u] == dfn[u] )
    {
        sccnum++;
        while( 1 )
        {
            int tmp = s.top();
            s.pop();
            vis[tmp] = 0;
            scc[tmp] = sccnum;
            if( low[tmp] == dfn[tmp] )
                break;
        }
    }
}

int main()
{
    while(~scanf("%d%d", &n, &m))
    {
        init();
        for( int i = 1; i <= n; i++ )
            scanf("%d", &pay[i]);
        int u, v;
        for( int i = 1; i <= m; i++ )
        {
            scanf("%d%d", &u, &v);
            p[i].st = u;
            p[i].ed = v;
            add(u, v);
        }
        for( int i = 1; i <= n; i++ )
        {
            if( !dfn[i] )
                tarjan(i);
        }
        for( int i = 1 ; i <= m ; i++ )
        {
            if( scc[ p[i].st ] != scc[ p[i].ed ] )
            //如果起点终点不在同一个联通块里，起点所在联通块出度++ ，终点的入度++ 
            {
                in[ scc[p[i].ed] ] ++;
            }
        }
        for( int i = 1; i <= n; i++ )
            vis[i] = inf;
        int ans1 = 0, ans2 = 0;
        for( int i = 1; i <= sccnum; i++ )
        {
            if (!in[i])
            {
            	ans1++;
            }
        }
        for( int i = 1; i <= n; i++ )
        {
            int tmp = scc[i];
            if( in[tmp] == 0 )
            {
                vis[tmp] = min(vis[tmp], pay[i]);
            }
        }
        for( int i = 1; i <= sccnum; i++ )
        {
            if( vis[i] != inf )
                ans2 += vis[i];
        }
        printf("%d %d\n", ans1, ans2);
    }
    return 0;
}

/*

12 16
1 2 3 4 5 6 7 8 9 10 11 12
1 3
3 2
2 1
3 4
2 4
3 5
5 4
4 6
6 4
7 4
7 12
7 8
8 7
8 9
10 9
11 10


*/