[csu/coj 1083]贪心

题意：给定n个线段，问能不能把x,y,z个长度为1,2,3的线段不重合地放进去。

思路：首先如果n个线段长度比要放的长度之和小，则无解，否则先考虑放2和3，如果2和3放下了1肯定可以放下（这是显然的）。于是我们贪心先把n个线段放满长度为3的线段，然后再考虑删去长度为3的线段来放长度为2的线段，删的时候要选择删去以后空闲的线段长度最多的删，比如某个线段本身有1的空闲线段，这时如果删去一条放在上面的长度为3的线段，则空闲线段变为4，这种情况优先删，其它情况次之。实现上采用优先队列，保存每个线段的长度为3的线段个数和空闲线段长度（只有0和1两种可能），然后按照前面说的优先级重载小于号就ok了，非常方便。

 #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 <stack>
 #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
 #define mp(a, b) make_pair(a, b)
 #define pb(a) push_back(a)

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

 const int dx[] = {, , -, , , , -, -};
 const int dy[] = {-, , , , , -, , - };
 const int maxn = 4e5 + ;
 const int md = 1e9 + ;
 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 cnt3, rest;
     bool operator < (const Node &that) const {
         return that.rest && that.cnt3 || that.cnt3 > cnt3 && !(rest && cnt3);
     }
     constructInt2(Node, cnt3, rest);
 };

 pii node[maxn];
 int a[maxn], tot, s, t, x, y, z, m;
 int Left[maxn], Right[maxn];
 int main() {
     //freopen("in.txt", "r", stdin);
     while (cin >> s >> t) {
         cin >> x >> y >> z;
         cin >> m;
         rep_up0(i, m) {
             sint2(node[i].first, node[i].second);
         }
         sort(node, node + m);
         tot = ;
         int R = -;
         int cs = ;
         rep_up0(i, m) {
             if (node[i].first - R > ) {
                 Left[cs] = R + ;
                 Right[cs ++] = node[i].first - ;
             }
             max_update(R, node[i].second);
         }
         if (R < 1e9) {
             Left[cs] = R + ;
             Right[cs ++] = 1e9;
         }
         rep_up0(i, cs) {
             int L = max(s, Left[i]), R = min(t, Right[i]);
             if (R >= L) a[tot ++] = R - L + ;
         }
         sort(a, a + tot);
         //rep_up0(i, tot) cout << a[i] << " ";
         int sum = ;
         rep_up0(i, tot) sum += a[i];
         if (sum < x +  * y +  * z) {
             puts("NO");
             continue;
         }
         priority_queue<Node> Q;
         int cnt2 = ;
         rep_up0(i, tot) {
             Q.push(Node(a[i] / , a[i] %  % ));
             cnt2 += a[i] %  / ;
         }

         while (!Q.empty()) {
             if (cnt2 >= y) break;
             Node Hnode = Q.top(); Q.pop();
             if (Hnode.rest) {
                 if (Hnode.cnt3) {
                     cnt2 += ;
                     Q.push(Node(Hnode.cnt3 - , ));
                 }
             }
             else {
                 if (Hnode.cnt3) {
                     cnt2 ++;
                     Q.push(Node(Hnode.cnt3 - , ));
                 }
             }
         }

         int cnt3 = ;
         while (!Q.empty()) {
             Node Hnode = Q.top(); Q.pop();
             cnt3 += Hnode.cnt3;
         }

         puts(cnt2 >= y && cnt3 >= z? "YES" : "NO") ;
     }
     return ;
 }