HDU 1827 — Summer Holiday

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

思路：记录环中权值最小的数，计算入度为0的强连通分量；

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>

using namespace std;
#define inf 999999999;
const int N = 1100;
const int M = 2100;
int head[N];
int DFN[N], low[N], stack[N]; 
int vis[N], a[N], val[N];
int taj, edgenum, top, time;
int belong[N];
int n, m;
int in[N];

struct node
{
    int from, to, nex;
}edge[M];

void add(int u, int v)
{
    edge[edgenum].from = u;
    edge[edgenum].to = v;
    edge[edgenum].nex = head[u];
    head[u] = edgenum++;
}

void tarjan(int u)
{
    DFN[u] = low[u] = time++;
    vis[u] = 1;
    stack[top++] = u;
    for(int i = head[u];i!=-1;i = edge[i].nex)
    {
        int v = edge[i].to;
        if(DFN[v] == -1)
        {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if(vis[v])
        low[u] = min(low[u], DFN[v]);
    }
    if(DFN[u] == low[u])
    {
        taj++;
        val[taj] = a[u];
        while(1)
        {
            int now = stack[--top];
            vis[now] = 0;
            belong[now] = taj;
            val[taj] = min(val[taj], a[now]);
            if(now == u)
            break;
        }
    }
}

void init()
{
    memset(head, -1, sizeof(head));
    memset(DFN, -1, sizeof(DFN));
    memset(low, -1, sizeof(low));
    memset(vis, 0, sizeof(vis));
    top = 0, taj = 0, time = 0;
    edgenum = 0;
}

int main()
{
    while(scanf("%d%d", &n, &m)!=EOF)
    {
        init();
        for(int i = 1;i<=n;i++)
        scanf("%d", &a[i]);
        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] == -1)
            tarjan(i);
        }
        memset(in, 0, sizeof(in));
        for(int i = 0;i<edgenum;i++)
        {
            int u = edge[i].from;
            int v = edge[i].to;
            if(belong[u]!=belong[v])
            in[belong[v]]++;
        }
        int ans = 0, sum = 0;
        for(int i = 1;i<=taj;i++)
        {
            if(in[i] == 0)
            {
                sum+=val[i];
                ans++;
            }
        }
        printf("%d %d\n", ans, sum);
    }
    return 0;
}