ZOJ-3261-Connections in Galaxy War [逆向并查集]

题目传送门

题意:

每个星星都有自己的power,星星之间有隧道可以联通，星星只能向比自己power更多的星星求救，如果最大的power相同就向序号更小的星星求救。

有两种操作:

“query x” 询问x可以求救的星星序号

“destroy x y” 破坏x - y连接的通道

思路:

如果直接对并查集进行删除操作会很麻烦，所以就将所有操作全部存起来逆向操作，先将所有没有被破坏的通道合并，然后逆向询问，破坏即为建边合并。

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <queue>
#include <map>
#include <set>
using namespace std;
int N, M, Q;
int val[12000];
map<int, map<int,int> >mark;
struct node1{
int x,y;
}P[22000];
struct node2{
char ch[10];
int x,y;
}K[50000+1000];
int fa[12000], ans[50000+1000];
void init()
{
for (int i = 0; i < 10000; i++)
{
fa[i]=i;
}
return ;
}
int find(int x)
{
while (x!=fa[x]) x = fa[x];
return x;
}
void unite(int x, int y)
{
int a = find(x);
int b = find(y);
if (a!=b)
{
if (val[a]==val[b])
{
if (a<b)
fa[b] = a;
else
fa[a] = b;
}
else
{
if (val[a]>val[b])
fa[b] = a;
else
fa[a] = b;
}
}
}
int main(void)
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);

bool flag = false;
while (~scanf("%d", &N))
{
init();
for (int i = 0; i < N; i++)
scanf("%d", &val[i]);
scanf("%d", &M);
for (int i = 0; i < M; i++)
scanf("%d %d", &P[i].x, &P[i].y);
scanf("%d", &Q);
for (int i = 0; i < Q; i++)
{
scanf(" %s", K[i].ch);
if (K[i].ch[0]=='q')
scanf("%d", &K[i].x);
else
{
scanf("%d %d", &K[i].x, &K[i].y);
mark[K[i].x][K[i].y] = mark[K[i].y][K[i].x] = 1;
}
}
for (int i = 0; i < M; i++)
{
if (mark[P[i].x][P[i].y]==1)
continue;
unite(P[i].x, P[i].y);
}
int cnt = 0;
for (int i = Q-1; i>=0; i--)
{
if (K[i].ch[0]=='q')
{
int a = find(K[i].x);
if (a==K[i].x)
ans[cnt++] = -1;
else
{
if (val[a]>val[K[i].x])
ans[cnt++]=a;
else
ans[cnt++]=-1;
}
}
else
{
unite(K[i].x, K[i].y);
}
}
if (flag)
printf("\n");
else
flag = true;
for (int i = cnt-1; i>=0; i--)
printf("%d\n", ans[i]);
mark.clear();
}
return 0;
}