luogu1197 [JSOI2008]星球大战

题目大意

　　有一个无向图，每次删除一个节点，求删除后图中连通块的个数。（如果两个星球可以通过现存的以太通道直接或间接地连通，则这两个星球在同一个连通块中）

题解

　　连通块？用并查集可以找到一个连通块，但是并查集不支持删除呀！所以我们将删点改为造点并连边就可以用并查集解决这个问题了。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int MAX_EDGE = 200010, MAX_NODE = MAX_EDGE * 2;
int TotNode, TotEdge;

struct Node;
struct Edge;

struct Edge
{
    Node *To;
    Edge *Next;
}_edges[MAX_EDGE * 2];
int _eCount;

struct Node
{
    Node *Father;
    Edge *Head;
    int AddRank;
    bool Built;
}_nodes[MAX_NODE], *AddOrder[MAX_NODE];
int Ans[MAX_NODE];

void AddEdge(Node *from, Node *to)
{
    Edge *e = _edges + _eCount++;
    e->To = to;
    e->Next = from->Head;
    from->Head = e;
}

void Build(int uId, int vId)
{
    AddEdge(_nodes + uId, _nodes + vId);
    AddEdge(_nodes + vId, _nodes + uId);
}

Node *FindRoot(Node *cur)
{
    return cur->Father == cur ? cur : cur->Father = FindRoot(cur->Father);
}

void Init(int n)
{
    for (int i = 0; i < n; i++)
        _nodes[i].Father = _nodes + i;
}

bool Join(Node *a, Node *b)
{
    Node *root1 = FindRoot(a), *root2 = FindRoot(b);
    if (root1 == root2)
        return false;
    else
    {
        root1->Father = root2;
        return true;
    }
}

bool Cmp(Node *a, Node *b)
{
    return a->AddRank < b->AddRank;
}

int main()
{
    scanf("%d%d", &TotNode, &TotEdge);
    for (int i = 0; i < TotEdge; i++)
    {
        int u, v;
        scanf("%d%d", &u, &v);
        Build(u, v);
    }
    int attackCnt;
    scanf("%d", &attackCnt);
    for (int i = 0; i < attackCnt; i++)
    {
        int v;
        scanf("%d", &v);
        _nodes[v].AddRank = attackCnt - i;
    }
    for (int i = 0; i < TotNode; i++)
        AddOrder[i] = _nodes + i;
    sort(AddOrder, AddOrder + TotNode, Cmp);
    int setCnt = 0;
    Init(TotNode);
    for (int i = 0; i < TotNode; i++)
    {
        AddOrder[i]->Built = true;
        setCnt++;
        for (Edge *e = AddOrder[i]->Head; e; e = e->Next)
            if (e->To->Built)
                setCnt -= Join(AddOrder[i], e->To);
        Ans[i]=setCnt;
    }
    for (int i = TotNode - 1; i >= TotNode - attackCnt - 1; i--)
        printf("%d\n", Ans[i]);
    return 0;
}