loj 1412(树上最长直径的应用)

题目链接：http://lightoj.com/volume_showproblem.php?problem=1412

思路：好久没写题解了，有点手生，这题从昨天晚上wa到现在终于是过了。。。思想其实很简单，就是预处理出每一块的最长直径，然后每次询问的时候直接查询就可以了。

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

const int MAXN = (100000 + 100);
typedef pair<int,int>Pair;

vector<int >g[MAXN];
vector<Pair >blocks;
int n,m,q,len,_count,max_len;
bool mark[MAXN];

int dfs(int u,int father)
{
    if(!mark[u])mark[u] = true, _count ++;
    int first = 0, second = 0;
    for(int i = 0; i < (int)g[u].size(); i ++) {
        int v = g[u][i];
        if(v == father)continue;
        int tmp = dfs(g[u][i],u) + 1;
        if(tmp > first)second = first, first = tmp;
        else if(tmp > second)second = tmp;
    }
    if(first + second > len)len = first + second;
    return first;
}

int cmp(Pair p, Pair q)
{
    if(p.second != q.second){
        return p.second > q.second;
    }else
        return p.first > q.first;
}

int main()
{
    int u,v,_case,t=1;
    scanf("%d",&_case);
    while(_case--){
        scanf("%d%d",&n,&m);
        for(int i=0;i<=n;i++)g[i].clear();
        while(m--){
            scanf("%d%d",&u,&v);
            g[u].push_back(v);
            g[v].push_back(u);
        }
        len=0;
        max_len = 0;
        memset(mark,false,sizeof(mark));
        blocks.clear();
        for(int i=1;i<=n;i++){
            if(!mark[i]){
                len=0;
                _count = 0;
                dfs(i,-1);
                blocks.push_back(make_pair(len,_count));
                max_len = max(max_len,len);
            }
        }
        sort(blocks.begin(),blocks.end(),cmp);
        scanf("%d",&q);
        printf("Case %d:
",t++);
        while(q--){
            int k;
            scanf("%d",&k);
            if(k > blocks[0].second){
                puts("impossible");
            }else if(k <= max_len + 1){
                printf("%d
",k - 1);
            }else {
                int ans = (1 << 30);
                for(int i=0; i<(int)blocks.size(); i++){
                    if(k > blocks[i].second)break;
                    ans = min (ans, (blocks[i].first)+2*(k-blocks[i].first-1));
                }
                printf("%d
",ans);
            }
        }
                            
    }
    return 0;
}