[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\n", 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 }