hdu2767&&hdu3836 Proving Equivalences（Tarjan+缩点）

http://acm.hdu.edu.cn/showproblem.php?pid=2767

题意：给你n个命题，m个推导，求增加多少条推导就能使命题两两等价。

ps：这里终于看懂九野巨的模板了，刚开始就是边的输入方式不懂，怎么模拟都不对。后来一想，这尼玛是头插法啊，我就说怎么似曾相识。看来我数据结构水的一比啊。也好，让自己复习下数据结构。

思路：命题转化为节点，推导转化为边，求再增加多少条边就可以将其变为强连通图。不同于上一道题判断是否为连通图，这题求的是添加的边数。这里采用了把强连通分量转化为缩点，这样就变成了有向无环图。连接有向无环图中入度为0与出度为0的点，相当于入度为0的有root个点，出度为0的有leaf个点，统计数量。因为肯定有剩余的，而剩余的乱连就行，而乱连也不会影响其连通性，所以要求最大值。关于缩点详解，看的这篇博客。主要就是将其染色，不同颜色的累加到需要连接的数组中，便于统计。

#include <stdio.h>
#include <algorithm>
#include <stdlib.h>
#include <string.h>
#include <iostream>
#include <stack>
#include <vector>

using namespace std;

typedef long long LL;

const int N = 100010;
const int INF = 1e8;

stack<int>S;

int dfn[N], low[N], head[N], Belong[N], in[N], out[N], countt, time, n, m, pos;
bool instack[N];

struct Edge
{
    int to, next;
}edge[N];

void add(int u, int v)
{
    edge[pos].to = v;
    edge[pos].next = head[u];
    head[u] = pos++;
}

void init()
{
    time = countt = pos = 0;
    memset(in, 0, sizeof(in));
    memset(out, 0, sizeof(out));
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(head, -1, sizeof(head));
    memset(instack, false, sizeof(instack));
}

void Tarjan(int u)
{
    int v;
    dfn[u] = low[u] = ++time;
    instack[u] = true;
    S.push(u);
    for(int i = head[u]; i != -1; i = edge[i].next)
    {
        v = edge[i].to;
        if(dfn[v] == 0)
        {
            Tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if(instack[v] == 1)
        {
            low[u] = min(low[u], dfn[v]);
        }
    }
    if(dfn[u] == low[u])
    {
        countt ++;
        do
        {
            v = S.top();
            S.pop();
            instack[v] = false;
            Belong[v] = countt;
        }while(u != v);
    }
}

int main()
{
   // freopen("in.txt", "r", stdin);
    int t, u, v;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &n, &m);
        init();
        for(int i = 1; i <= m; i++)
        {
            scanf("%d%d", &u, &v);
            add(u, v);
        }
        //缩点
        for(int i = 1; i <= n; i++)
        {
            if(dfn[i] == 0)
                Tarjan(i);
        }
        if(countt == 1)
        {
            printf("0\n");
            continue;
        }
        for(int i = 1; i <= n; i++)
        {
            for(int k = head[i]; k != -1; k = edge[k].next)
            {
                int j = edge[k].to;
                if(Belong[i] != Belong[j])
                {
                    out[Belong[i]]++;
                    in[Belong[j]]++;
                }
            }
        }
        //找叶子和根度为0的数量
        int root = 0, leaf = 0;
        for(int i = 1; i <= countt; i++)
        {
            if(in[i] == 0) root++;
            if(out[i] == 0) leaf++;
        }
        printf("%d\n", max(root, leaf));
    }
    return 0;
}

顺便贴上hdu3836，代码几乎一毛一样。

#include <stdio.h>
#include <algorithm>
#include <stdlib.h>
#include <string.h>
#include <iostream>
#include <stack>
#include <vector>

using namespace std;

typedef long long LL;

const int N = 100010;
const int INF = 1e8;

stack<int>S;

int dfn[N], low[N], head[N], Belong[N], in[N], out[N], countt, time, n, m, pos;
bool instack[N];

struct Edge
{
    int to, next;
}edge[N];

void add(int u, int v)
{
    edge[pos].to = v;
    edge[pos].next = head[u];
    head[u] = pos++;
}

void init()
{
    time = countt = pos = 0;
    memset(in, 0, sizeof(in));
    memset(out, 0, sizeof(out));
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(head, -1, sizeof(head));
    memset(instack, false, sizeof(instack));
}

void Tarjan(int u)
{
    int v;
    dfn[u] = low[u] = ++time;
    instack[u] = true;
    S.push(u);
    for(int i = head[u]; i != -1; i = edge[i].next)
    {
        v = edge[i].to;
        if(dfn[v] == 0)
        {
            Tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if(instack[v] == 1)
        {
            low[u] = min(low[u], dfn[v]);
        }
    }
    if(dfn[u] == low[u])
    {
        countt ++;
        do
        {
            v = S.top();
            S.pop();
            instack[v] = false;
            Belong[v] = countt;
        }while(u != v);
    }
}

int main()
{
   // freopen("in.txt", "r", stdin);
    int u, v;
    while(~scanf("%d%d", &n, &m))
    {
        init();
        for(int i = 1; i <= m; i++)
        {
            scanf("%d%d", &u, &v);
            add(u, v);
        }
        //缩点
        for(int i = 1; i <= n; i++)
        {
            if(dfn[i] == 0)
                Tarjan(i);
        }
        if(countt == 1)
        {
            printf("0\n");
        }
        else
        {
            for(int i = 1; i <= n; i++)
            {
                for(int k = head[i]; k != -1; k = edge[k].next)
                {
                    int j = edge[k].to;
                    if(Belong[i] != Belong[j])
                    {
                        out[Belong[i]]++;
                        in[Belong[j]]++;
                    }
                }
            }
            //找叶子和根度为0的数量
            int root = 0, leaf = 0;
            for(int i = 1; i <= countt; i++)
            {
                if(in[i] == 0) root++;
                if(out[i] == 0) leaf++;
            }
            printf("%d\n", max(root, leaf));
        }
    }
    return 0;
}