hdu

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

缩点后,统计入度为0的点有多少个,那么这些点都是需要被通知的,但是这些点可能也是被缩的,所以每次在这个点所属集合找一个最小值即可.

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define ll long long
#define inf 0x7f7f7f7f
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)

#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d
", x)
#define lowbit(x)   (x)&(-x)
#define Read()  freopen("a.txt", "r", stdin)
#define Write() freopen("dout.txt", "w", stdout);

using namespace std;
#define N 1100
//N为最大点数
#define M 2100
//M为最大边数
int n, m;//n m 为点数和边数

struct Edge{
    int from, to, nex;
    bool sign;//是否为桥
}edge[M<<1];
int head[N], edgenum;
void add(int u, int v){//边的起点和终点
    Edge E={u, v, head[u], false};
    edge[edgenum] = E;
    head[u] = edgenum++;
}

int DFN[N], Low[N], Stack[N], top, Time; //Low[u]是点集{u点及以u点为根的子树} 中(所有反向弧)能指向的(离根最近的祖先v) 的DFN[v]值(即v点时间戳)
int taj;//连通分支标号，从1开始
int Belong[N];//Belong[i] 表示i点属于的连通分支
bool Instack[N];
vector<int> bcc[N]; //标号从1开始 每个强连通分量包含原图中的点

void tarjan(int u ,int fa){
    DFN[u] = Low[u] = ++ Time ;
    Stack[top ++ ] = u ;
    Instack[u] = 1 ;

    for (int i = head[u] ; ~i ; i = edge[i].nex ){
        int v = edge[i].to ;
        if(DFN[v] == -1)
        {
            tarjan(v , u) ;
            Low[u] = min(Low[u] ,Low[v]) ;
            if(DFN[u] < Low[v])
            {
                edge[i].sign = 1;//为割桥
            }
        }
        else if(Instack[v]) Low[u] = min(Low[u] ,DFN[v]) ;
    }
    if(Low[u] == DFN[u]){
        int now;
        taj ++ ; bcc[taj].clear();
        do{
            now = Stack[-- top] ;
            Instack[now] = 0 ;
            Belong [now] = taj ;
            bcc[taj].push_back(now);
        }while(now != u) ;
    }
}

void tarjan_init(int all){
    memset(DFN, -1, sizeof(DFN));
    memset(Instack, 0, sizeof(Instack));
    top = Time = taj = 0;
    for(int i=1;i<=all;i++)if(DFN[i]==-1 )tarjan(i, i); //注意开始点标！！！
}
vector<int>G[N];
int du[N];
void suodian(){
    memset(du, 0, sizeof(du));
    for(int i = 1; i <= taj; i++)G[i].clear();
    for(int i = 0; i < edgenum; i++){
        int u = Belong[edge[i].from], v = Belong[edge[i].to];
        if(u!=v)
        {
            G[u].push_back(v), du[v]++;
           // printf("%d %d
",u,v);
        }
    }
}
void init(){memset(head, -1, sizeof(head)); edgenum=0;}
int p[N];
int main()
{
    //Read();
    int a,b;
    while(~scanf("%d%d",&n,&m))
    {
        init();
        for(int i=1;i<=n;i++) scanf("%d",&p[i]);
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&a,&b);
            add(a,b);
        }
        tarjan_init(n);
        suodian();
        int x=0,ans=0,y;
        for(int i=1;i<=taj;++i)
        {
            y=1<<30;
            if(du[i]==0) //出度为0点的个数
            {
                x++;
                for(int j=0;j<bcc[i].size();++j)
                    y=min(y,p[bcc[i][j]]);
                ans+=y;
            }
        }
        //printf("%d
",j);
        printf("%d %d
",x,ans);
    }
    return 0;
}