ZOJ 3261 - Connections in Galaxy War ，并查集删边

In order to strengthen the defense ability, many stars in galaxy allied together and built many bidirectional tunnels to exchange messages. However, when the Galaxy War began, some tunnels were destroyed by the monsters from another dimension. Then many problems were raised when some of the stars wanted to seek help from the others.

In the galaxy, the stars are numbered from 0 to N-1 and their power was marked by a non-negative integer p_i. When the star A wanted to seek help, it would send the message to the star with the largest power which was connected with star A directly or indirectly. In addition, this star should be more powerful than the star A. If there were more than one star which had the same largest power, then the one with the smallest serial number was chosen. And therefore, sometimes star A couldn't find such star for help.

Given the information of the war and the queries about some particular stars, for each query, please find out whether this star could seek another star for help and which star should be chosen.

Input

There are no more than 20 cases. Process to the end of file.

For each cases, the first line contains an integer N (1 <= N <= 10000), which is the number of stars. The second line contains N integers p₀, p₁, ... , p_n-1 (0 <= p_i <= 1000000000), representing the power of the i-th star. Then the third line is a single integer M (0 <= M<= 20000), that is the number of tunnels built before the war. Then M lines follows. Each line has two integers a, b (0 <= a, b <= N - 1, a != b), which means star a and star b has a connection tunnel. It's guaranteed that each connection will only be described once.

In the (M + 2)-th line is an integer Q (0 <= Q <= 50000) which is the number of the information and queries. In the following Q lines, each line will be written in one of next two formats.

"destroy a b" - the connection between star a and star b was destroyed by the monsters. It's guaranteed that the connection between star a and star b was available before the monsters' attack.

"query a" - star a wanted to know which star it should turn to for help

There is a blank line between consecutive cases.

Output

For each query in the input, if there is no star that star a can turn to for help, then output "-1"; otherwise, output the serial number of the chosen star.

Print a blank line between consecutive cases.

Sample Input

2
10 20
1
0 1
5
query 0
query 1
destroy 0 1
query 0
query 1

Sample Output

1
-1
-1
-1

 大致题意：

　　n个星球，每个星球有一定的power值，某些星球是直接或间接相连的；

　　当某个星球想求助时会找到相连的里面的power值最大而且大于自己的一个星球；

　　先在给定这些power值并给定两两相连的信息，然后又q个操作，destroy a b是删除a b直接相连的边（保证存在）；

　　query a求向谁求助，如果不能求助输出-1 

解题思路：

　　并查集删边操作，直接删边困难，那就改成先把不用删的边连起来；

　　全部信息保存下来，再逆序处理，这样子删边就变成了加边，就很好操作了，同样访问也是没有问题的；

 　　（字体略小建议复制下来看T^T）

 1 #include <cstdio>
 2 #include <map>
 3 using namespace std;
 4 const int N=10005;
 5 struct link{ int a,b,flag;}t[N<<1],q[N<<3];
 6 int f[N],pw[N],ans[N<<3]; char c[10]; map<int,bool>mark;
 7 int sf(int x) {return x==f[x]?x:f[x]=sf(f[x]);}
 8 void Union(int a,int b){
 9     int fa=sf(a),fb=sf(b);
10     if(fa!=fb) if(pw[fa]>pw[fb] || pw[fa]==pw[fb]&&fa<fb) f[fb]=fa; else f[fa]=fb;
11 }
12 int main(){
13     bool flag=0; int n,m,x,y,Q;
14     while(~scanf("%d",&n)){  if(flag) puts(""); flag=1; 
15         mark.clear();
16         for(int i=0;i<n;i++) f[i]=i,scanf("%d",&pw[i]);
17         scanf("%d",&m);
18         for(int i=0;i<m;i++){
19             scanf("%d%d",&t[i].a,&t[i].b);
20             if(t[i].a>t[i].b) swap(t[i].a,t[i].b);//前小后大 
21         } scanf("%d",&Q);
22         for(int i=0;i<Q;i++){
23             scanf("%s",&c);
24             if(c[0]=='d'){
25                 scanf("%d%d",&q[i].a,&q[i].b);
26                 if(q[i].a>q[i].b) swap(q[i].a,q[i].b);//前小后大 
27                 q[i].flag=1; mark[q[i].a*N+q[i].b]=1; //标记区分
28             } else scanf("%d",&q[i]),q[i].flag=0;//标记区分
29         } for(int i=0;i<m;i++) if(!mark[t[i].a*N+t[i].b]) Union(t[i].a,t[i].b);//不用破坏的先连起来 
30         for(int i=Q-1;i>=0;i--){//从后往前处理 
31             if(q[i].flag) Union(q[i].a,q[i].b); 
32             else {
33                 int fa=sf(q[i].a); 
34                 ans[i]=pw[fa]>pw[q[i].a]?fa:-1;
35             }
36         } for(int i=0;i<Q;i++) if(!q[i].flag) printf("%d\n",ans[i]);
37     } return 0;
38 }

 

转载于:https://www.cnblogs.com/nicetomeetu/p/5165031.html