题意:给定 n 个工作的最好开始时间,和持续时间,现在有两种方法,第一种,如果当前的工作能够恰好在最好时间开始,那么就开始,第二种,如果不能,那么就从前找最小的时间点,来完成。
析:直接暴力,每次都先去看看能不能在最好时间完成,如果不能,就去找最小的时间点。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const LL LNF = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e2 + 100; const int mod = 1e9 + 7; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline int gcd(int a, int b){ return b ? gcd(b, a%b) : a; } inline int lcm(int a, int b){ return a * b / gcd(a, b); } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } struct Node{ int l, r; Node() { } Node(int ll, int rr) : l(ll), r(rr) { } bool operator < (const Node &p) const{ return l < p.l; } }; Node a[205]; int main(){ while(scanf("%d", &n) == 1){ int s, d; int cnt = 1; a[0] = Node(0, 0); cin >> s >> d; printf("%d %d\n", s, s+d-1); a[cnt++] = Node(s, s+d-1); for(int i = 1; i < n; ++i){ cin >> s >> d; --d; sort(a, a+cnt); if(a[cnt-1].r < s){ printf("%d %d\n", s, s+d); a[cnt++] = Node(s, s+d); continue; } bool ok = false; for(int j = 0; j < cnt-1; ++j){ if(s > a[j].r && s+d < a[j+1].l){ ok = true; printf("%d %d\n", s, s+d); a[cnt++] = Node(s, s+d); break; } } if(!ok){ bool ok1 = false; for(int j = 1; j < cnt; ++j){ if(a[j].l - a[j-1].r > d+1){ printf("%d %d\n", a[j-1].r+1, a[j-1].r+d+1); a[cnt++] = Node(a[j-1].r+1, a[j-1].r+d+1); ok1 = true; break; } } if(!ok1){ printf("%d %d\n", a[cnt-1].r+1, a[cnt-1].r+1+d); a[cnt] = Node(a[cnt-1].r+1, a[cnt-1].r+d+1); ++cnt; } } } } return 0; }