[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。

 #pragma comment(linker, "/STACK:10240000,10240000")

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstdlib>
 #include <cstring>
 #include <map>
 #include <queue>
 #include <deque>
 #include <cmath>
 #include <vector>
 #include <ctime>
 #include <cctype>
 #include <set>
 #include <bitset>
 #include <functional>
 #include <numeric>
 #include <stdexcept>
 #include <utility>

 using namespace std;

 #define mem0(a) memset(a, 0, sizeof(a))
 #define mem_1(a) memset(a, -1, sizeof(a))
 #define lson l, m, rt << 1
 #define rson m + 1, r, rt << 1 | 1
 #define define_m int m = (l + r) >> 1
 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
 #define rep_down1(a, b) for (int a = b; a > 0; a--)
 #define all(a) (a).begin(), (a).end()
 #define lowbit(x) ((x) & (-(x)))
 #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) {}
 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 #define pchr(a) putchar(a)
 #define pstr(a) printf("%s", a)
 #define sstr(a) scanf("%s", a)
 #define sint(a) scanf("%d", &a)
 #define sint2(a, b) scanf("%d%d", &a, &b)
 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
 #define pint(a) printf("%d\n", a)
 #define test_print1(a) cout << "var1 = " << a << endl
 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl

 typedef long long LL;
 typedef pair<int, int> pii;
 typedef vector<int> vi;

 const int dx[] = {, , -, , , , -, -};
 const int dy[] = {-, , , , , -, , - };
 const int maxn = 3e4 + ;
 const int md = ;
 const int inf = 1e9 + ;
 const LL inf_L = 1e18 + ;
 const double pi = acos(-1.0);
 const double eps = 1e-;

 template<class T>T gcd(T a, T b){return b==?a:gcd(b,a%b);}
 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
 template<class T>T condition(bool f, T a, T b){return f?a:b;}
 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
 int make_id(int x, int y, int n) { return x * n + y; }

 struct Node {
     int r, s, a, d;
     Node(int r = , int s = , int a = , int d = ): r(r), s(s), a(a), d(d) {}
     bool operator < (const Node that) const {
         return r < that.r;
     }
 };
 Node node[];
 bool occ[];

 int main() {
     //freopen("in.txt", "r", stdin);
     int n;
     while (cin >> n) {
         int maxt = ;
         bool ok = true;
         mem0(occ);
         rep_up0(i, n) {
             int s, t, a, d, tmp = ;
             cin >> s >> t >> a >> d;
             node[i] = Node(t, s, a, d);
             max_update(maxt, t);
         }
         sort(node, node + n);
         rep_up0(i, n) {
             int cur = node[i].r;
             while (node[i].s < ) {
                 while (occ[cur] && cur) cur--;
                 if (!cur || node[i].a <= ) {
                     ok = false;
                     break;
                 }
                 occ[cur] = true;
                 node[i].s += node[i].a;
                 node[i].a -= node[i].d;
             }
             if (!ok) break;
         }
         rep_down1(i, maxt) {
             if (!occ[i]) {
                 int s = , p;
                 rep_up0(j, n) {
                     if (node[j].r >= i) {
                         int dat = node[j].a;
                         if (node[j].s + dat > ) dat =  - node[j].s;
                         if (dat > s) {
                             s = dat;
                             p = j;
                         }
                     }
                 }
                 if (s) {
                     node[p].s += s;
                     node[p].a -= node[p].d;
                 }
             }
         }
         int ans = ;
         rep_up0(i, n) {
             ans += node[i].s;
         }
         if (ok) cout << ans << endl;
         else puts("you are unlucky");
     }
 }