Gym 100851E Easy Problemset (水题，模拟)

Posted 2020-08-09 dwtfukgv

题意：给定 n 个裁判，然后每个都一些题目，现在要从每一个按顺序去选出 k 个题，并且这 k 个要按不递减顺序，如果没有，就用50补充。

析：就按他说的来，直接模拟就好。

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 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 LL LNF = 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, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"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;
}
struct node{
    int val, id;
    node(int v, int i) : val(v), id(i) { }
};
int a[15][15];

int main(){
    freopen("easy.in", "r", stdin);
    freopen("easy.out", "w", stdout);
    int k;
    while(scanf("%d %d", &n, &k) == 2){
        int x = 0;
        memset(a, -1, sizeof a);
        for(int i = 0; i < n; ++i){
            scanf("%d", &m);
            x = Max(x, m);
            for(int j = 0; j < m; ++j)
                scanf("%d", a[i]+j);
        }

        int cnt = 0, ans = 0;
        for(int i = 0; i < x && cnt < k; ++i){
            for(int j = 0; j < n && cnt < k; ++j)
                if(a[j][i] >= ans){
                    ans += a[j][i];
                    ++cnt;
                }
                else if(a[j][i] == -1) { ans += 50; ++cnt;  break; }
        }

        while(cnt < k)  ans += 50, ++cnt;
        printf("%d\n", ans);
    }
    return 0;
}