国庆出游

代码链接：https://hihocoder.com/problemset/problem/1041
参考链接：

1. http://blog.csdn.net/u010535824/article/details/40652865
2. http://blog.csdn.net/u013076044/article/details/45795939

思路：
（1）从根节点开始利用bitset求出每个节点的所有子孙节点；
（2）对于长度为total的目标序列seq，从根节点开始深搜。

• 若当前节点等于seq[cur_index],则++cur_index；
• 若cur_index==total，置ok=1,返回；
• 当cur_index小于total时，对当前节点的每一个满足条件的子孙节点进行深搜（条件：该子节点可达seq[cur_index]&&未曾访问）；
• 若深搜后cur_index大小未变说明没有可达序列，函数返回。

#include<iostream>#include<vector>#include<bitset>#include <stdio.h>#include <string.h>using namespace std;const int limited = 101;vector<int> u[limited];vector<int> seq(limited,0);int no_visited[limited][limited];bitset<limited> reach[limited];int ok = 0, total = 0,cur_index=0;void can_reach(int v) {    reach[v][v] = 1;    for (int i = 0;i < u[v].size();++i) {        int child = u[v][i];        can_reach(child);        reach[v] |= reach[child];    }}int judge(int v) {    if (v == seq[cur_index])        ++cur_index;    if (cur_index >= total) {        ok = 1;        return 0;    }    while (cur_index < total) {        int mark = cur_index,c=seq[cur_index];        for (int i = 0;i < u[v].size();++i) {            int node = u[v][i];            if (reach[node][c] && no_visited[v][node]) {                no_visited[v][node] = 0;                judge(node);            }        }        if (mark == cur_index)            return 1;    }}int main() {    int test = 0;    scanf("%d", &test);    memset(no_visited, 0, sizeof(no_visited));    while (test--) {        ok = 0;cur_index = 0;        for (int i = 0;i < limited;++i) {            u[i].clear();            reach[i].reset();        }        int n,f,c;        scanf("%d", &n);        for (int i = 1;i < n;++i) {            scanf("%d%d", &f, &c);            u[f].push_back(c);            no_visited[f][c] = 1;        }        scanf("%d", &total);        for (int i = 0;i < total;++i)            scanf("%d", &seq[i]);        can_reach(1);        judge(1);        printf("%s\n", ok == 1 ? "YES" : "NO");    }    return 0;}