CodeForces 732D Exams (二分)

时间:2023-03-08 21:50:22
CodeForces 732D Exams (二分)

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;
}