题目大意：
给定一个N个点、M条边的无向图Graph，以及从点1开始进行DFS形成的树Tree，定义"T-Simple Circle"为Graph中的环，要求其中只含一条不属于Tree的边。
将Graph中的一些边进行染色，使得其中每个T-simple Circle都至少包含一条被染色的边，求最少需要染色的边数。
N≤2e3，M≤2e4
本题关键的一点在于Tree是一棵DFS生成树，这样Tree以外的边只可能将某个点与它在Tree中的祖先相连（用反证法可以证明，只有这样才能维持DFS树的性质）。
也就是说，每条Tree以外的边都相当于在DFS生成树上划定了一条深度单调递增（递减）的链，问题转化为：最少染色多少条边，可以使每条链上都至少有一条边被染色。
不难发现，对Tree以外的边进行染色的覆盖效率远小于对Tree上的边进行染色，因此只需考虑DFS生成树的边。
类比直线上的区间选点问题，本题也可以用类似的贪心思路。
题解：http://blog.csdn.net/sd_invol/article/details/9963741
直线上区间选点问题的证明：http://blog.csdn.net/dgq8211/article/details/7534776
C++11代码（貌似HDOJ可以交C++11？）：
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #include <vector>
5 #include <functional>
6
7 const int maxN = 2000 + 5;
8 const int maxM = 20000 + 5;
9
10 std::vector<int> toVec[maxN];
11 int father[maxN]; //father in the DFS tree
12 int depth[maxN]; //depth in the DFS tree
13 bool covered[maxN]; //whether the edge (x - father[x]) is covered (used in greedy algorithm)
14
15 struct Range
16 {
17 int head, tail; //We guarantee that depth[head] >= depth[tail]
18
19 void swapEndPoint()
20 {
21 if (depth[head] < depth[tail])
22 std::swap(head, tail);
23 }
24 bool operator < (const Range& rhs) const
25 {
26 return depth[tail] > depth[rhs.tail] ||
27 (depth[tail] == depth[rhs.tail] && depth[head] < depth[rhs.head]);
28 //high depth -> 0 --- 0 --- 0 --- 0 --- 0 -> low depth
29 //greater: x --------- x
30 //less: x --------------------- x
31 }
32 };
33
34 Range range[maxM];
35 int N, M;
36
37 void init()
38 {
39 memset(father, 0, sizeof(father));
40 memset(depth, 0, sizeof(depth));
41 memset(covered, 0, sizeof(covered));
42 for (int i = 1; i <= N; i++)
43 toVec[i].clear();
44 }
45
46 /// @brief swap head and tail so that depth[head] >= depth[tail]
47 /// this function is called after depth[] is set, before sorting ranges for greedy algorithm
48 void initRange()
49 {
50 for (int i = 0; i < M - N + 1; i++)
51 range[i].swapEndPoint();
52 }
53
54 bool input()
55 {
56 scanf("%d%d", &N, &M);
57 if (N == 0)
58 return false;
59
60 init();
61 for (int u, v, i = 0; i < N - 1; i++) //(N - 1) Edges in DFS tree
62 {
63 scanf("%d%d", &u, &v);
64 toVec[u].push_back(v);
65 toVec[v].push_back(u);
66 }
67 for (int u, v, i = N - 1; i < M; i++)
68 {
69 scanf("%d%d", &u, &v);
70 range[i - N + 1] = {u, v}; //The end points may be swapped later
71 }
72
73 return true;
74 }
75
76 ///@brief DFS process, setting depth[] and father[]
77 void dfs(int cur, int last)
78 {
79 father[cur] = last;
80 depth[cur] = depth[last] + 1;
81 for (auto to: toVec[cur])
82 {
83 if (to == last)
84 continue;
85 dfs(to, cur);
86 }
87 }
88
89 ///@brief wrapper of DFS function
90 void setDepthAndFather()
91 {
92 depth[0] = 0;
93 dfs(1, 0);
94 }
95
96 int solve()
97 {
98 setDepthAndFather();
99 initRange();
100 std::sort(range, range + M - N + 1); //(M - N + 1) Edges that does not belong to the DFS tree
101
102 ///@return last if edge (last, father[last]) should be covered
103 /// 0 if no edge should be covered in this chain
104 auto getCoverEdge = [] (const Range& rg) -> int
105 {
106 int last = rg.head;
107
108 //higher depth -> head -> tail -> lower depth
109 for (int cur = rg.head; cur != rg.tail; cur = father[cur])
110 {
111 if (covered[cur])
112 return 0;
113 last = cur;
114 }
115 return last;
116 };
117
118 // ///@debug
119 // for (int i = 1; i <= N; i++)
120 // printf("father[%d] = %d, depth[%d] = %d\n", i, father[i], i, depth[i]);
121
122 int ans = 0;
123 for (int i = 0; i < M - N + 1; i++)
124 {
125 int coverId = getCoverEdge(range[i]);
126 if (coverId == 0)
127 continue;
128 ans += 1;
129 covered[coverId] = true;
130 }
131
132 return ans;
133 }
134
135 int main()
136 {
137 while (input())
138 printf("%d\n", solve());
139 return 0;
140 }
原文地址：https://www.cnblogs.com/Onlynagesha/p/8448560.html
时间： 2024-10-11 08:17:01