题意:某人要考试,有n天考m个科目,然后有m个科目要考试的时间和要复习多少天才能做,问你他最早考完所有科目是什么时间。
析:二分答案,然后在判断时,直接就是倒着判,很明显后出来的优先,也就是一个栈。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #define debug() puts("++++"); #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e5 + 5; const int mod = 1e9 + 7; const int dr[] = {-1, 1, 0, 0}; const int dc[] = {0, 0, 1, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int a[maxn], b[maxn]; int ok[maxn], cnt[maxn]; bool judge(int mid){ memset(ok, -1, sizeof ok); int idx = 0, num = 0; for(int i = mid; i > 0; --i){ if(a[i] && ok[a[i]] == -1) cnt[++idx] = a[i], ++ok[a[i]], ++num; else{ while(idx && !b[cnt[idx]]) --idx; if(!idx) continue; ++ok[cnt[idx]]; if(ok[cnt[idx]] == b[cnt[idx]]) --idx; } } return num == m && !idx; } int solve(){ int l = m, r = n+1; while(l < r){ int mid = (l + r) >> 1; if(judge(mid)) r = mid; else l = mid + 1; } return l == n+1 ? -1 : l; } int main(){ while(scanf("%d %d", &n, &m) == 2){ for(int i = 1; i <= n; ++i) scanf("%d", a+i); a[n+1] = 0; b[0] = -1; for(int i = 1; i <= m; ++i) scanf("%d", b+i); printf("%d\n", solve()); } return 0; }