标签：the kings problem   hdu 3861   强连通缩点   最小路径覆盖   

The King’s Problem

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3 2
1 2
1 3

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题意：n个城市m条有向边，把这些城市分成若干个州，分的原则是（1）u和v可以互相到达的话他们两个必须在同一个州（2）同一个州里任意两个城市u和v要满足u可以到达v或者v可以到达u。问州的最小个数是多少。

思路：先用Tarjan算法进行缩点，在缩点后的图上进行二分图匹配，最后求得最小路径覆盖=强连通个数-最大匹配数。

可以看一下：

http://blog.csdn.net/hellobabygogo3/article/details/7900812

http://www.cnblogs.com/ka200812/archive/2011/07/31/2122641.html

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

#define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 1005;
const int MAXN = 5005;
const int MAXM = 100010;

int n,m;

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

int head[MAXN],tot;
int Low[MAXN],DFN[MAXN],Belong[MAXN],Stack[MAXN];
int Index,scc,top;
bool Inq[MAXN];

int uN,vN;
int linker[MAXN];
bool used[MAXN];
int hed[MAXN],num;

void init()
{
    tot=0;num=0;
    memset(hed,-1,sizeof(hed));
    memset(head,-1,sizeof(head));
}

void add(int u,int v)
{
    e[num].to=v;
    e[num].next=hed[u];
    hed[u]=num++;
}

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

void Tarjan(int u)
{
    int v;
    Low[u]=DFN[u]=++Index;
    Stack[top++]=u;
    Inq[u]=true;
    for (int i=head[u];i+1;i=edge[i].next)
    {
        v=edge[i].to;
        if (!DFN[v])
        {
            Tarjan(v);
            if (Low[u]>Low[v]) Low[u]=Low[v];
        }
        else if (Inq[v]&&Low[u]>DFN[v])
            Low[u]=DFN[v];
    }
    if (Low[u]==DFN[u])
    {
        scc++;
        do{
            v=Stack[--top];
            Inq[v]=false;
            Belong[v]=scc;
        }while (v!=u);
    }
}
void solve(int N)
{
    memset(DFN,0,sizeof(DFN));
    memset(Inq,false,sizeof(Inq));
    Index=scc=top=0;
    for (int i=1;i<=N;i++)
        if (!DFN[i])
            Tarjan(i);
    for (int u=1;u<=N;u++)
    {
        for (int i=head[u];~i;i=edge[i].next)
        {
            int v=edge[i].to;
            if (Belong[u]==Belong[v]) continue;
            add(Belong[u],Belong[v]);
        }
    }
    uN=vN=scc;
}

bool dfs(int u)
{
    for (int i=hed[u];~i;i=e[i].next)
    {
        int v=e[i].to;
        if (!used[v])
        {
            used[v]=true;
            if (linker[v]==-1||dfs(linker[v]))
            {
                linker[v]=u;
                return true;
            }
        }
    }
    return false;
}

int hungary()
{
    int ans=0;
    memset(linker,-1,sizeof(linker));
    for (int i=1;i<=uN;i++)
    {
        memset(used,false,sizeof(used));
        if (dfs(i)) ans++;
        if (ans==uN) break;
    }
    return ans;
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
    int i,j,t,u,v;
    sf(t);
    while (t--)
    {
        sff(n,m);
        init();
        for (i=0;i<m;i++)
        {
            sff(u,v);
            addedge(u,v);
        }
        solve(n);
        int ans=hungary();
        printf("scc=%d \n",scc);
        printf("%d\n",scc-ans);
    }
    return 0;
}

The King’s Problem (hdu 3861 强连通缩点+最小路径覆盖)

标签：the kings problem   hdu 3861   强连通缩点   最小路径覆盖   

原文地址：http://blog.csdn.net/u014422052/article/details/47205335