[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。
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 <stack> 16 #include <set> 17 #include <bitset> 18 #include <functional> 19 #include <numeric> 20 #include <stdexcept> 21 #include <utility> 22 23 using namespace std; 24 25 #define mem0(a) memset(a, 0, sizeof(a)) 26 #define mem_1(a) memset(a, -1, sizeof(a)) 27 #define lson l, m, rt << 1 28 #define rson m + 1, r, rt << 1 | 1 29 #define define_m int m = (l + r) >> 1 30 #define rep_up0(a, b) for (int a = 0; a < (b); a++) 31 #define rep_up1(a, b) for (int a = 1; a <= (b); a++) 32 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--) 33 #define rep_down1(a, b) for (int a = b; a > 0; a--) 34 #define all(a) (a).begin(), (a).end() 35 #define lowbit(x) ((x) & (-(x))) 36 #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) {} 37 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 38 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 39 #define pchr(a) putchar(a) 40 #define pstr(a) printf("%s", a) 41 #define sstr(a) scanf("%s", a) 42 #define sint(a) scanf("%d", &a) 43 #define sint2(a, b) scanf("%d%d", &a, &b) 44 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c) 45 #define pint(a) printf("%d\n", a) 46 #define test_print1(a) cout << "var1 = " << a << endl 47 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl 48 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl 49 #define mp(a, b) make_pair(a, b) 50 #define pb(a) push_back(a) 51 52 typedef long long LL; 53 typedef pair<int, int> pii; 54 typedef vector<int> vi; 55 56 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1}; 57 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 }; 58 const int maxn = 4e5 + 7; 59 const int md = 1e9 + 7; 60 const int inf = 1e9 + 7; 61 const LL inf_L = 1e18 + 7; 62 const double pi = acos(-1.0); 63 const double eps = 1e-6; 64 65 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);} 66 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;} 67 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;} 68 template<class T>T condition(bool f, T a, T b){return f?a:b;} 69 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];} 70 int make_id(int x, int y, int n) { return x * n + y; } 71 72 struct Node { 73 int cnt3, rest; 74 bool operator < (const Node &that) const { 75 return that.rest && that.cnt3 || that.cnt3 > cnt3 && !(rest && cnt3); 76 } 77 constructInt2(Node, cnt3, rest); 78 }; 79 80 pii node[maxn]; 81 int a[maxn], tot, s, t, x, y, z, m; 82 int Left[maxn], Right[maxn]; 83 int main() { 84 //freopen("in.txt", "r", stdin); 85 while (cin >> s >> t) { 86 cin >> x >> y >> z; 87 cin >> m; 88 rep_up0(i, m) { 89 sint2(node[i].first, node[i].second); 90 } 91 sort(node, node + m); 92 tot = 0; 93 int R = -1; 94 int cs = 0; 95 rep_up0(i, m) { 96 if (node[i].first - R > 1) { 97 Left[cs] = R + 1; 98 Right[cs ++] = node[i].first - 1; 99 } 100 max_update(R, node[i].second); 101 } 102 if (R < 1e9) { 103 Left[cs] = R + 1; 104 Right[cs ++] = 1e9; 105 } 106 rep_up0(i, cs) { 107 int L = max(s, Left[i]), R = min(t, Right[i]); 108 if (R >= L) a[tot ++] = R - L + 1; 109 } 110 sort(a, a + tot); 111 //rep_up0(i, tot) cout << a[i] << " "; 112 int sum = 0; 113 rep_up0(i, tot) sum += a[i]; 114 if (sum < x + 2 * y + 3 * z) { 115 puts("NO"); 116 continue; 117 } 118 priority_queue<Node> Q; 119 int cnt2 = 0; 120 rep_up0(i, tot) { 121 Q.push(Node(a[i] / 3, a[i] % 3 % 2)); 122 cnt2 += a[i] % 3 / 2; 123 } 124 125 while (!Q.empty()) { 126 if (cnt2 >= y) break; 127 Node Hnode = Q.top(); Q.pop(); 128 if (Hnode.rest) { 129 if (Hnode.cnt3) { 130 cnt2 += 2; 131 Q.push(Node(Hnode.cnt3 - 1, 0)); 132 } 133 } 134 else { 135 if (Hnode.cnt3) { 136 cnt2 ++; 137 Q.push(Node(Hnode.cnt3 - 1, 1)); 138 } 139 } 140 } 141 142 int cnt3 = 0; 143 while (!Q.empty()) { 144 Node Hnode = Q.top(); Q.pop(); 145 cnt3 += Hnode.cnt3; 146 } 147 148 puts(cnt2 >= y && cnt3 >= z? "YES" : "NO") ; 149 } 150 return 0; 151 }