题意： 给出一些点，每个点有权值，然后有一些边，相连。无向的。然后有一些操作
query a.表示从a出发的能到达的所有点权值最大的点的编号（相同取编号最小，而且权值要比自己大）
destory a,b 表示删除连接a,b的边
思路：一开始就按题目写...写着写着发现不对劲啊..好像难搞...然后搜了一下，思路很巧妙，正难则反的思路，先把操作，图都保存下来，然后从后往前查询，根节点保存权值最大的点，如果是destroy相当于连接两个点，起始图就是经过查询后保留下来的子图
#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define maxn 10005
#define LL long long
int cas=1,T;
int pre[10005];
int vis[10005];
int p[maxn];
typedef struct edge
{
int v;
bool del;
}edge;
typedef struct
{
int ord;
int u,v;
}order;
order o[50005];
vector<edge> city[maxn];
int ans[50005];
int Find(int x)
{
int r = x;
while (r!=pre[r])
r=pre[r];
int i = x,j;
while (pre[i]!=r)
{
j=pre[i];
pre[i]=r;
i=j;
}
return r;
}
void mix(int x,int y)
{
int fx = Find(x),fy=Find(y);
if (p[fx]>p[fy] || p[fx]==p[fy]&&fx<fy)
{
pre[fy]=fx;
}
else
pre[fx]=fy;
}
int main()
{
int n;
int first=0;
while (scanf("%d",&n)!=EOF)
{
if (first)
printf("\n");
first=1;
for (int i =0;i<n;i++)
{
pre[i]=i;
}
for (int i = 0;i<n;i++)
{
scanf("%d",&p[i]);
city[i].clear();
}
int m;
scanf("%d",&m);
edge temp;
temp.del=0;
for (int i = 0;i<m;i++)
{
int a,b;
scanf("%d%d",&a,&b);
temp.v=b;
city[a].push_back(temp);
temp.v=a;
city[b].push_back(temp);
// mix(a,b);
}
int q;
scanf("%d",&q);
for (int i = 0;i<q;i++)
{
int a,b;
char s[10];
scanf("%s",s);
if (s[0]=='q')
{
scanf("%d",&a);
o[i].ord=2;
o[i].u=a;
}
if (s[0]=='d')
{
scanf("%d%d",&a,&b);
o[i].ord=1;
o[i].u=a;
o[i].v=b;
int j;
for (j=0;j<city[a].size();j++)
if (city[a][j].v == b)
break;
city[a][j].del = 1;
for (j=0;j<city[b].size();j++)
if (city[b][j].v==a)
break;
city[b][j].del=1;
}
}
for (int i=0;i<n;i++)
{
for (int j = 0;j<city[i].size();j++)
{
if (city[i][j].del == 0)
mix(i,city[i][j].v);
}
}
int pos = 0;
for (int i = q-1;i>=0;i--)
{
if (o[i].ord==1)
mix(o[i].u,o[i].v);
else
{
int x = Find(o[i].u);
if (p[x]!=p[o[i].u])
ans[pos++]=x;
else
ans[pos++]=-1;
}
}
for (int i = pos-1;i>=0;i--)
printf("%d\n",ans[i]);
}
//freopen("in","r",stdin);
//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
return 0;
} 题目
Description
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 pi. 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 integersp0, p1, ... , pn-1 (0 <= pi <= 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
