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  • [csu1603]贪心

    题意:有n门考试,对于考试i,不复习它能考si分,复习它的每小时能提高ai分,每过一小时ai会减小di,也就是说,连续复习某门科目每小时提高的分为ai, ai-di, ai-2di...,但每门考试最高分不超过100分,给定每门考试的考试时间,且考试本身不占时间,合理安排复习计划,问能否让所有考试都不低于60分,如果可以,总和最大为多少。

    思路:比较明显的贪心,但难以想到正确的贪心策略。我们把问题分成两步,所有考试全部通过60分和得分总和最大。对于第一步,对于不复习分数低于60分的科目,先复习一段时间让它分数达到60分,贪心从考试时间往前延伸找空闲时间复习(一旦找不到空闲时间,则无解了)。对于第二步,按时间顺序的逆序处理,对于某个时间t,枚举所有考试时间大于等于t的科目,找到某一科目,使得复习它提高的分数最高,那么时间t就用来复习这门功课。

      1 #pragma comment(linker, "/STACK:10240000,10240000")
      2 
      3 #include <iostream>
      4 #include <cstdio>
      5 #include <algorithm>
      6 #include <cstdlib>
      7 #include <cstring>
      8 #include <map>
      9 #include <queue>
     10 #include <deque>
     11 #include <cmath>
     12 #include <vector>
     13 #include <ctime>
     14 #include <cctype>
     15 #include <set>
     16 #include <bitset>
     17 #include <functional>
     18 #include <numeric>
     19 #include <stdexcept>
     20 #include <utility>
     21 
     22 using namespace std;
     23 
     24 #define mem0(a) memset(a, 0, sizeof(a))
     25 #define mem_1(a) memset(a, -1, sizeof(a))
     26 #define lson l, m, rt << 1
     27 #define rson m + 1, r, rt << 1 | 1
     28 #define define_m int m = (l + r) >> 1
     29 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
     30 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
     31 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
     32 #define rep_down1(a, b) for (int a = b; a > 0; a--)
     33 #define all(a) (a).begin(), (a).end()
     34 #define lowbit(x) ((x) & (-(x)))
     35 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
     36 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
     37 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
     38 #define pchr(a) putchar(a)
     39 #define pstr(a) printf("%s", a)
     40 #define sstr(a) scanf("%s", a)
     41 #define sint(a) scanf("%d", &a)
     42 #define sint2(a, b) scanf("%d%d", &a, &b)
     43 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
     44 #define pint(a) printf("%d
    ", a)
     45 #define test_print1(a) cout << "var1 = " << a << endl
     46 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
     47 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
     48 
     49 typedef long long LL;
     50 typedef pair<int, int> pii;
     51 typedef vector<int> vi;
     52 
     53 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
     54 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
     55 const int maxn = 3e4 + 7;
     56 const int md = 10007;
     57 const int inf = 1e9 + 7;
     58 const LL inf_L = 1e18 + 7;
     59 const double pi = acos(-1.0);
     60 const double eps = 1e-4;
     61 
     62 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
     63 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
     64 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
     65 template<class T>T condition(bool f, T a, T b){return f?a:b;}
     66 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
     67 int make_id(int x, int y, int n) { return x * n + y; }
     68 
     69 struct Node {
     70     int r, s, a, d;
     71     Node(int r = 0, int s = 0, int a = 0, int d = 0): r(r), s(s), a(a), d(d) {}
     72     bool operator < (const Node that) const {
     73         return r < that.r;
     74     }
     75 };
     76 Node node[100];
     77 bool occ[1000];
     78 
     79 int main() {
     80     //freopen("in.txt", "r", stdin);
     81     int n;
     82     while (cin >> n) {
     83         int maxt = 0;
     84         bool ok = true;
     85         mem0(occ);
     86         rep_up0(i, n) {
     87             int s, t, a, d, tmp = 0;
     88             cin >> s >> t >> a >> d;
     89             node[i] = Node(t, s, a, d);
     90             max_update(maxt, t);
     91         }
     92         sort(node, node + n);
     93         rep_up0(i, n) {
     94             int cur = node[i].r;
     95             while (node[i].s < 60) {
     96                 while (occ[cur] && cur) cur--;
     97                 if (!cur || node[i].a <= 0) {
     98                     ok = false;
     99                     break;
    100                 }
    101                 occ[cur] = true;
    102                 node[i].s += node[i].a;
    103                 node[i].a -= node[i].d;
    104             }
    105             if (!ok) break;
    106         }
    107         rep_down1(i, maxt) {
    108             if (!occ[i]) {
    109                 int s = 0, p;
    110                 rep_up0(j, n) {
    111                     if (node[j].r >= i) {
    112                         int dat = node[j].a;
    113                         if (node[j].s + dat > 100) dat = 100 - node[j].s;
    114                         if (dat > s) {
    115                             s = dat;
    116                             p = j;
    117                         }
    118                     }
    119                 }
    120                 if (s) {
    121                     node[p].s += s;
    122                     node[p].a -= node[p].d;
    123                 }
    124             }
    125         }
    126         int ans = 0;
    127         rep_up0(i, n) {
    128             ans += node[i].s;
    129         }
    130         if (ok) cout << ans << endl;
    131         else puts("you are unlucky");
    132     }
    133 }
    View Code
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  • 原文地址:https://www.cnblogs.com/jklongint/p/4474442.html
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