ZOJ 3261 Connections in Galaxy War

Connections in Galaxy War

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Time Limit: 3 Seconds      Memory Limit: 32768 KB

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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 integerp_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 starA directly or indirectly. In addition, this star should be more powerful than the starA. 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 starA 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 containsN 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 integerM (0 <= M <= 20000), that is the number of tunnels built before the war. ThenM lines follows. Each line has two integers a, b (0 <=a, b <= N - 1, a != b), which means stara 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 followingQ lines, each line will be written in one of next two formats.

• "destroy a b" - the connection between star a and starb was destroyed by the monsters. It's guaranteed that the connection between stara 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

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Author: MO, Luyi

Source: ZOJ Monthly, November 2009

转载地址：http://blog.youkuaiyun.com/ggggiqnypgjg/article/details/6621481

虽然转载过来，但是加上了自己的注释。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<vector>
#define N 50005

using namespace std;

int pre[N];
int num[N];
int a[N],b[N],ans[N];
char str[20];
vector<int> relation[N];

int find(int x)
{
    if(x!=pre[x])
       pre[x]=find(pre[x]);
    return pre[x];//找老大
}

void Union(int x,int y)
{
    int fx = find(x);
    int fy = find(y);
    if(fx!=fy)
    {
        if(num[fx]>num[fy])
            pre[fy] = fx;//如果fx的能量比fy的能量，那么使得fy的父节点指向fx
        else if(num[fy]>num[fx])
            pre[fx] = fy;//和上述相反
        else if(num[fx]==num[fy])
            {
                if(fx<fy)
                   pre[fy] = fx;
                else
                   pre[fx] = fy;
            }
    }
}

int main()
{
    int i,x,y,t,T,M,Q,l,cnt;
    scanf("%d",&T);
    vector<int>::iterator flag;
    while(1)
    {
        for(i=0;i<T;i++)
           scanf("%d",&num[i]);
        scanf("%d",&M);
        for(i=0;i<M;i++)
        {
            scanf("%d%d",&x,&y);
            if(x>y) swap(x,y);
            relation[x].push_back(y);//把通道的关系存下来
        }
        scanf("%d",&Q);
        for(i=0;i<Q;i++)
        {
            scanf("%s",str);
            if(str[0]=='q')
            {
                scanf("%d",&b[i]);
                a[i]=-1;//标记下q操作
            }
            else
            {
                scanf("%d%d",&a[i],&b[i]);
                if(a[i]>b[i])
                    swap(a[i],b[i]);
                for(flag=relation[a[i]].begin();flag!=relation[a[i]].end();flag++)
                    if(*flag==b[i])
                       *flag=-1;//判断删除操作，然后把删除通道的操作标记好
            }
        }

        for(i=0;i<=T;i++)
             pre[i]=i;//初始化
        for(i=0;i<T;i++)
        {
            for(flag=relation[i].begin();flag!=relation[i].end();flag++)
            {
                if(*flag!=-1)
                    Union(i,*flag);
               // printf("%d %d..\n", i, *flag);

            }//如果不是删除的操作，就并起来
        }
        l=0;
        for(i=Q-1;i>=0;i--)//关键在于从后往前遍历操作，就等于把删除看成了加边
        {
            if(a[i]==-1)
            {
                    cnt=find(b[i]);
                    if(num[cnt]!=num[b[i]])
                       ans[l++]=cnt;
                    else
                       ans[l++]=-1;
            }//查询操作
            else
                    Union(a[i],b[i]);//删除操作
        }
        for(i=l-1;i>=0;i--)
                printf("%d\n",ans[i]);
        for(i=0;i<T;i++)
                relation[i].clear();
        if(scanf("%d", &T)!=EOF)
            printf("\n");
        else
            break;

    }
    return 0;
}