Round 310(Div.1) B. Case of Fugitive

Posted 2020-07-08

Round 310(Div.1) B. Case of Fugitive

Andrewid the android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost fully covered with water.

The only dry land there is an archipelago of n narrow islands located in a row. For more comfort let‘s represent them as non-intersecting segments on a straight line: island i has coordinates [li,?ri], besides, r_i?<?l_i+1 for 1?≤?i?≤?n?-?1.

To reach the goal, Andrewid needs to place a bridge between each pair of adjacent islands. A bridge of length a can be placed between the i-th and the (i?+?1)-th islads, if there are such coordinates of x and y, that li?≤?x?≤?ri, l_i?+?1?≤?y?≤?r_i?+?1 and y?-?x?=?a.

The detective was supplied with m bridges, each bridge can be used at most once. Help him determine whether the bridges he got are enough to connect each pair of adjacent islands.

1. #include <stdio.h> 

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2. #include <string.h> 

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3.  

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4. #include <queue> 

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5. #include <vector> 

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6. #include <algorithm> 

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7.  

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8. const int N = 200000 + 5; 

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9.  

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10. long long left[N]; 

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11. long long right[N]; 

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12.  

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13. struct segment_t 

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14. { 

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15. long long left; 

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16. long long right; 

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17. int index; 

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18. } segment[N]; 

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19.  

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20. struct left_order 

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21. { 

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22. bool operator() (const segment_t& first, const segment_t& second) 

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23. { return first.left < second.left; } 

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24. }; 

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25.  

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26. struct right_order 

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27. { 

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28. bool operator() (const segment_t& first, const segment_t& second) 

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29. { 

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30. if (first.right != second.right) 

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31. return first.right > second.right; 

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32. return first.index < second.index; 

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33. } 

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34. }; 

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35.  

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36. struct bridge_t 

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37. { 

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38. long long length; 

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39. int index; 

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40. } bridge[N]; 

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41.  

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42. struct length_order 

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43. { 

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44. bool operator() (const bridge_t& first, const bridge_t& second) 

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45. { return first.length < second.length; } 

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46. }; 

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47.  

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48. int answer[N]; 

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49.  

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50. int main() 

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51. { 

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52. #ifndef ONLINE_JUDGE 

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53. freopen("input.txt", "r", stdin); 

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54. // freopen("input.txt", "w", stdout); 

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55. #endif // ONLINE_JUDGE 

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56. int n, m; 

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57. scanf("%d %d", &n, &m); 

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58. for (int i = 1; i <= n; ++i) { 

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59. scanf("%I64d %I64d", &left[i], &right[i]); 

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60. } 

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61. for (int i = 1; i <= m; ++i) { 

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62. scanf("%I64d", &bridge[i].length); 

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63. bridge[i].index = i; 

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64. } 

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65. for (int i = 1; i < n; ++i) { 

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66. segment[i].left = left[i+1] - right[i]; 

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67. segment[i].right = right[i+1] - left[i]; 

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68. segment[i].index = i; 

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69. } 

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70. std::sort(segment+1, segment+n, left_order()); 

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71. std::sort(bridge+1, bridge+1+m, length_order()); 

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72. std::priority_queue<segment_t, std::vector<segment_t>, right_order> pq; 

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73. int ptr = 0; 

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74. int assigned_count = 0; 

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75. for (int i = 1; i <= m; ++i) { 

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76. for (; ptr < n && segment[ptr].left <= bridge[i].length; ptr++) { 

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77. pq.push(segment[ptr]); 

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78. } 

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79. for (; !pq.empty() && pq.top().right < bridge[i].length; pq.pop()); 

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80. if (pq.empty()) { 

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81. continue; 

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82. } 

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83. assigned_count += 1; 

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84. answer[pq.top().index] = bridge[i].index; 

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85. pq.pop(); 

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86. } 

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87. if (assigned_count == n - 1) { 

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88. puts("Yes"); 

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89. for (int i = 1; i < n; ++i) { 

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90. printf("%d ", answer[i]); 

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91. } 

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92. printf("\n"); 

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93. } 

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94. else puts("No"); 

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95. return 0; 

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96. } 

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