[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
", 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[8] = {0, 0, -1, 1, 1, 1, -1, -1};
const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
const int maxn = 4e5 + 7;
const int md = 1e9 + 7;
const int inf = 1e9 + 7;
const LL inf_L = 1e18 + 7;
const double pi = acos(-1.0);
const double eps = 1e-6;

template<class T>T gcd(T a, T b){return b==0?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 = 0;
        int R = -1;
        int cs = 0;
        rep_up0(i, m) {
            if (node[i].first - R > 1) {
                Left[cs] = R + 1;
                Right[cs ++] = node[i].first - 1;
            }
            max_update(R, node[i].second);
        }
        if (R < 1e9) {
            Left[cs] = R + 1;
            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 + 1;
        }
        sort(a, a + tot);
        //rep_up0(i, tot) cout << a[i] << " ";
        int sum = 0;
        rep_up0(i, tot) sum += a[i];
        if (sum < x + 2 * y + 3 * z) {
            puts("NO");
            continue;
        }
        priority_queue<Node> Q;
        int cnt2 = 0;
        rep_up0(i, tot) {
            Q.push(Node(a[i] / 3, a[i] % 3 % 2));
            cnt2 += a[i] % 3 / 2;
        }

        while (!Q.empty()) {
            if (cnt2 >= y) break;
            Node Hnode = Q.top(); Q.pop();
            if (Hnode.rest) {
                if (Hnode.cnt3) {
                    cnt2 += 2;
                    Q.push(Node(Hnode.cnt3 - 1, 0));
                }
            }
            else {
                if (Hnode.cnt3) {
                    cnt2 ++;
                    Q.push(Node(Hnode.cnt3 - 1, 1));
                }
            }
        }

        int cnt3 = 0;
        while (!Q.empty()) {
            Node Hnode = Q.top(); Q.pop();
            cnt3 += Hnode.cnt3;
        }

        puts(cnt2 >= y && cnt3 >= z? "YES" : "NO") ;
    }
    return 0;
}