1622Trees on the level

哈哈哈哈,我又来啦,这次是一道关于树的题目,不过输入很阴间,我的思路就是先建树,然后判断树中节点是否全都有权

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1622

代码:

#include<iostream>
#include<string>
#include<algorithm>
#include<string.h>
#include<cmath>
#include<vector>
#include<queue>
#include<unordered_set>
using namespace std;
typedef long long ll;
#define md (l+r)>>1
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
const int maxn = 257;
int tree[maxn][2];
int val[maxn];
int tot;
int cnt;
vector<string>V[maxn];
bool flg;
void build(string& pos, int i, int w, int rt) {
    if (i == pos.size()) {
        if (val[rt])flg = 0;//如果该节点已经有权值了(题目说了权值全是postive的才可以这样用,不然就加一个vis数组)
        val[rt] = w;
        cnt++;//有权值的节点数目+1
        return;
    }
    if (pos[i] == 'L') {
        if (!tree[rt][0])tree[rt][0] = ++tot;
        build(pos, i + 1, w, tree[rt][0]);
    }
    else {
        if (!tree[rt][1])tree[rt][1] = ++tot;
        build(pos, i + 1, w, tree[rt][1]);
    }
}
void level_order(int rt) {//层序遍历没啥好说的
    queue<int>Q;
    Q.push(rt);
    bool h = 0;
    while (!Q.empty()) {
        int t = Q.front(); Q.pop();
        if (h)printf(" ");
        printf("%d", val[t]);
        h = 1;
        if (tree[t][0])Q.push(tree[t][0]);
        if (tree[t][1])Q.push(tree[t][1]);
    }
    printf("\n");
}
int num = 0;
int main() {
    //freopen("test.txt", "r", stdin);
    string m;
    while (cin >> m) {
        if (m == "()") {
            num++; continue;
        }
        V[num].push_back(m);
    }
    for (int i = 0; i < num; i++) {
        memset(tree, 0, sizeof(tree));//初始化操作一定要做好,我flg没初始化还去问人家,呜呜呜...
        memset(val, 0, sizeof(val));
        tot = 1;//树中节点个数
        cnt = 0;//有权值的节点个数
        flg = 1;//预设树是满足条件的
        for (auto& j : V[i]) {
            //对付阴间输入就要用正道的光...STL..
            string w = j.substr(1, j.find(',') - 1);
            int m = atoi(w.c_str());
            string path = j.substr(j.find(',') + 1);
            path.pop_back();//也可以修改上面的substr,但是我难得动脑壳,直接把')'弹出去
            build(path, 0, m, 1);
            if (!flg)break;
        }
        if (cnt != tot)flg = 0;//如果个数不相等说明树中有些节点还没有权值,即构不成完整的树
        if (flg) {
            level_order(1);
        }
        else {
            printf("not complete\n");
        }
    }
    return 0;
}