贪心 URAL 1303 Minimal Coverage

 

题目传送门

 1 /*  2  题意：最少需要多少条线段能覆盖[0, m]的长度  3  贪心：首先忽略被其他线段完全覆盖的线段，因为选取更长的更优  4  接着就是从p=0开始，以p点为标志，选取 (node[i].l <= p && p < node[i+1].l)  5  详细解释：http://www.cnblogs.com/freezhan/p/3219046.html  6 */  7 #include <cstdio>  8 #include <iostream>  9 #include <algorithm> 10 #include <cstring> 11 #include <cmath> 12 using namespace std; 13 14 const int MAXN = 1e6 + 10; 15 const int INF = 0x3f3f3f3f; 16 struct Node 17 { 18 int l, r; 19 }node[MAXN], ans[MAXN]; 20 21 bool cmp(Node x, Node y) 22 { 23 if (x.l == y.l) return x.r > y.r; 24 else return x.l < y.l; 25 } 26 27 int main(void) //URAL 1303 Minimal Coverage 28 { 29 //freopen ("R.in", "r", stdin); 30 31 int m; 32 while (scanf ("%d", &m) == 1) 33  { 34 int n = 0; int u, v; 35 while (scanf ("%d%d", &u, &v) == 2 && (u || v)) 36 node[++n].l = u, node[n].r = v; 37 sort (node+1, node+1+n, cmp); 38 39 int k = 1; 40 for (int i=2; i<=n; ++i) //cover 41  { 42 if (node[k].l < node[i].l && node[k].r < node[i].r) 43  { 44 node[++k] = node[i]; 45  } 46  } 47 48 n = k; int p = 0; int cnt = 0; 49 node[n+1].l = node[n+1].r = m + 1; 50 for (int i=1; i<=n; ++i) 51  { 52 if (node[i].l <= p && p < node[i+1].l) 53  { 54 ans[++cnt] = node[i]; p = node[i].r; 55  } 56 if (p >= m) 57  { 58 printf ("%d\n", cnt); 59 for (int i=1; i<=cnt; ++i) 60  { 61 printf ("%d %d\n", ans[i].l, ans[i].r); 62  } 63 break; 64  } 65  } 66 67 if (p < m) puts ("No solution"); 68  } 69 70 return 0; 71 } 72 73 /* 74 No solution 75 */