UVA - 1608 Non-boring sequences : 分治
2017-08-03 20:15
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题目点此跳转
思路
题目意思是如果一个序列的任意连续子序列中至少有一个只出现一次的元素,则称这个序列是不无聊(non-boring)的。输入一个n(n≤200000)个元素的序列A(各个元素均为10 9 以内的非负整数),判断它是不是不无聊的。分治法
在整个A[1,, n]里找到一个只出现一次的元素,记为A[p],然后以此为分治点, 分别对A[1,, p-1] 和A[p+1,, n]执行同样的操作,如果哪个步骤没有找到这个只出现一次的元素,则返回false.
而判断一个元素只出现一次的方法是, 借用map,预处理出每个元素上一个与它相同的元素出现的下标,并记录在两个数组里.
代码
int n, a[maxn], pre[maxn], nxt[maxn]; map<int, int> m; bool solve(int L, int R) { if(L == R) return 1; int k = -1; for(int i = 0; L+i <= R-i; ++i) { if(pre[L+i] < L && nxt[L+i] > R) { k = L+i; break; } if(pre[R-i] < L && nxt[R-i] > R) { k = R-i; break; } } if(k == -1) return 0; bool f1 = 1, f2 = 1; if(L < k) f1 = solve(L, k-1); if(k < R) f2 = solve(k+1, R); return f1&f2; } int main() { int t; cin >> t; while(t--) { scanf("%d", &n); m.clear(); bool ok = 0; for(int i = 0; i < n; ++i) scanf("%d", &a[i]); for(int i = 0; i < n; ++i) { if(!m.count(a[i])) pre[i] = -1; else pre[i] = m[a[i]]; m[a[i]] = i; } m.clear(); for(int i = n-1; i >= 0; --i) { if(!m.count(a[i])) nxt[i] = n; else nxt[i] = m[a[i]]; m[a[i]] = i; } if(solve(0, n-1)) printf("non-boring\n"); else printf("boring\n"); } return 0; }
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