2016 Multi-University Training Contest 4 1010 The All-purpose Zero
2016-07-29 09:27
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题目链接:点击打开链接
题目大意:找LIS其中有0的话,0可以代替任何数。
解题思路:找到每个0之前的LIS,0作为前一段LIS的最后一位+1,前面的序列尽可能的小,于是加上一个剪枝操作,当前面的序列出现0~n,就可以剪掉了。
代码:
题目大意:找LIS其中有0的话,0可以代替任何数。
解题思路:找到每个0之前的LIS,0作为前一段LIS的最后一位+1,前面的序列尽可能的小,于是加上一个剪枝操作,当前面的序列出现0~n,就可以剪掉了。
代码:
#include<iostream> #include<vector> #include<cmath> #include<algorithm> #include<ctime> #include "cstdio" #include "string" #include "string.h" #include "map" #include "bitset" using namespace std; const int INF = 0x3f3f3f3f; int a[100001]; int main() { int T, cas = 0; scanf("%d", &T); while(T--) { int n, x, ans = 0, pos; scanf("%d", &n); memset(a, INF, sizeof(a)); for (int i = 0; i < n; i++) { scanf("%d", &x); if (x) { pos = lower_bound(a, a + n, x) - a; a[pos] = x; } else { pos = lower_bound(a, a + n, INF) - a; for (int j = pos - 1; j >= 0; j--) { a[j + 1] = a[j] + 1; if (a[j] == j) break; } a[0] = 0; } ans = max(ans, pos+1); } printf("Case #%d: %d\n", ++cas, ans); } return 0; }
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