題意
題目我截圖下來(lái)了，我大致解釋下，
AOJ 0033 Ball
。有編號(hào)1到10共10個(gè)球，從上方丟下去，入口處可以選擇進(jìn)入左邊或者右邊，最后10個(gè)球全部落下去后如果左右兩側(cè)都是從小到大的順序，則輸出YES；否則輸出NO。<喎?http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vcD4NCjxwPjxpbWcgYWx0PQ=="這里寫圖片描述" src="http://www.2cto.com/uploadfile/Collfiles/20151213/20151213091427178.jpg" title="\" />代碼
一開始我先測(cè)試了一下自己理解的題意是不是對(duì)的：
<code class="hljs cpp">#include<iostream>#include<vector>using namespace std;int main() { vector<int>left; vector<int>right; vector<int>all; bool flag = true; int n; cin >> n; if (n == 0) return -1; for (int i = 0; i < n; i++) { for (int j = 0; j < 10; j++) { int temp; cin >> temp; all.push_back(temp); } } for (int i = 0; i < n; i++) { for (int j = 0; j < 10; j++) { if (left.size() > 0) { if (all[10 * i + j] > left[left.size() - 1]) { left.push_back(all[10 * i + j]); } else { if (right.size() > 0) { if (all[10 * i + j] > right[right.size() - 1]) right.push_back(all[10 * i + j]); else flag = false; } else { right.push_back(all[10 * i + j]); } } } else { left.push_back(all[10 * i + j]); } } if (flag) cout << "YES" << endl; else cout << "NO" << endl; flag = true; } return 0;}</int></int></int></vector></iostream></code>后來(lái)提交代碼居然錯(cuò)了，什么鬼！！我用題目中的用例測(cè)試是對(duì)的啊，還是沒有發(fā)現(xiàn)原因在哪……
因?yàn)橹李}意是要求用DFS，所以改改代碼，思路一樣，再試試：
<code class="hljs cpp">#include<stdio.h>#include<queue>using namespace std;bool flag = true;void solve(queue<int>left, queue<int>right, queue<int>all) { if (all.size() > 0) { if (left.size() > 0) { if (all.front() > left.back()) { left.push(all.front()); all.pop(); solve(left, right, all); } else { if (right.size() > 0) { if (all.front() > right.back()) { right.push(all.front()); all.pop(); solve(left, right, all); } else if(all.size() == 0){ } else { flag = false; } } else { right.push(all.front()); all.pop(); solve(left, right, all); } } } else { left.push(all.front()); all.pop(); solve(left, right, all); } }}int main() { int n; scanf("%d", &n); for (; n > 0; n--) { queue<int>all; queue<int>left; queue<int>right; for (int i = 0; i < 10; i++) { int temp; scanf("%d", &temp); all.push(temp); } solve(left, right, all); if (flag) printf("YES\n"); else printf("NO\n"); } return 0;}</int></int></int></int></int></int></queue></stdio.h></code>這次終于可以了，證明我的思路沒有問題的呀！
找了份代碼過來(lái)，變量挺多的：
<code class="hljs cpp">#include<iostream>#include<stack>#include<queue>using namespace std;int main() { stack<int>b, c; int a[10]; bool which[11]; int data[11]; int index; int n, m, A; int i, j; cin >> n; for (i = 0; i<n; cin="" for="" index="0;" j="0;">> m; a[j] = m; data[j + 1] = 0; } b.push(0); c.push(0); while (index >= 0) { A = a[index]; if (b.top() < A && (data[A] != 1 && data[A] != 3)) { b.push(A); data[A] += 1; which[A] = true; } else if (c.top() < A && (data[A] != 2 && data[A] != 3)) { c.push(A); data[A] += 2; which[A] = false; } else { index--; if (index < 0) { break; } else if (which[a[index]]) { b.pop(); } else { c.pop(); } continue; } index++; if (index > 9) { cout << "YES" << endl; break; } } if (index < 0) { cout << "NO" << endl; } }}</n;></int></queue></stack></iostream></code>