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分类: C/C++

2011-06-01 15:27:32

两种方法,一种是并查集,另一种是排序+二分+二分图判定


并查集的方法是最简单的,不论是空间还是时间,都是最优的,但是不太好想到

 

排序+二分+二分图判定,比较容易在第一思路想到,但是实现就不如并查集那么简单,而且我动态建立图,然后超时最后一个点……

 

并查集做法:


总共只有两个集合,想一下这样的逻辑关系,a不和b在一起,a不和c在一起,那么b和c一定在一起

 

有什么想法吗?没错,我们可以对每个罪犯设置一个集合,每个罪犯的补集设置一个集合,通过不断维护着相互间的制约关系来进行判断,这时我们就可以用并查集来做到这一点

 

因为题目要求是最大的冲突,所以先将冲突值排序,从大到小依次解决,什么时候遇到了冲突,说明最大的冲突值就是他,这个贪心方法就不用证明的了吧,很显然的

程序代码如下:

 

  1. #include <iostream>
  2. #include <algorithm>
  3. #include <cstdio>
  4. #include <cstring>
  5. #include <cstdlib>

  6. using namespace std;

  7. const int M = 100000;
  8. const int N = 20000;

  9. struct EDGE {
  10.         int a;
  11.         int b;
  12.         int v;
  13. } edge[M];

  14. int f[N * 2];

  15. bool Compare(const EDGE &a, const EDGE &b)
  16. {
  17.         return a.v < b.v;
  18. }

  19. int Find(int x)
  20. {
  21.         if (f[x] == x) {
  22.                 return f[x];
  23.         } else {
  24.                 return f[x] = Find(f[x]);
  25.         }
  26. }

  27. int main()
  28. {
  29.         int n, m;

  30.         ios::sync_with_stdio(false);

  31.         cin >> n;
  32.         cin >> m;
  33.         for (int i = 0; i < m; i++) {
  34.                 cin >> edge[i].a;
  35.                 cin >> edge[i].b;
  36.                 cin >> edge[i].v;
  37.         }

  38.         for (int i = 0; i < n * 2; i++) {
  39.                 f[i] = i;
  40.         }

  41.         sort(edge, edge + m, Compare);

  42.         for (int i = m - 1; i >= 0; i--) {
  43.                 int x = Find(edge[i].a);
  44.                 int y = Find(edge[i].b);

  45.                 if (x == y) {
  46.                         cout << edge[i].v << endl;

  47.                         return 0;
  48.                 }

  49.                 int xx = Find(edge[i].a + n);
  50.                 int yy = Find(edge[i].b + n);

  51.                 f[x] = yy;
  52.                 f[y] = xx;
  53.         }

  54.         cout << "0" << endl;

  55.         return 0;
  56. }

排序+二分+二分图判定做法:


首先,我承认自己这个方法没有AC,因为我动态建图,开销比较大,最后一个点超时,比并查集慢得多……

 

说一下大体思路吧,因为是求最大值最小的问题,所以可以采用二分的方法来查找最优值,当然,前提是将冲突值排序。

 

在二分的时候,选定一个[left_border,right_border]范围内的中间冲突值,然后将所有大于等于该冲突的两个罪犯之间连一条边,代表他们不能在一起

 

建图建好后,然后用BFS进行二分图判定,具体见下面代码中的BinaryGraphJudge(),这里需要注意一下,由于建好后的图可能不连通,所以需要额外加一个visited_vertex变量来记录已经扩展了多少个点,如果在还没有扩展完的情况下队列里面就已经空了,那么就再选取一个没有被扩展过的点加入队列,继续进行BFS,直到visited_vertex == n就可以退出了

 

我感觉这里的二分是一个挑战,至少对我来说,可能大牛们对此已经不屑了,但是我感觉边界问题很难处理,详情可以看一下《编程珠玑》里的第四章——编写正确的程序

 

在下面这个程序里,在进行二分的时候,我能够保证的就是[right_border, m]里的所有冲突值都是可以避免的,[0, left_border)里的冲突值都是不能避免的(注意区间的开闭),最后,当left_border == right_border时,我可以确定left_border - 1就是我所需要的那个满足要求的冲突值最大的罪犯的编号

 

这里说一下people[0]的问题,输入的数据全部存储在people[1] ~ people[m]中,而这个people[0]就是可能的无解情况,虽然这个题里的数据没有0的情况,但我觉得应该考虑一下,所以,将people[0]的冲突值设为0,这样,当将题目中的关系全部满足后,仍然是个二分图的话,那么结果将是left_border == right_border == 1(不考虑语法问题…),所以将返回0,最后输出的也就是0

 

程序代码如下:

  1. #include <cstdio>
  2. #include <cstring>
  3. #include <cstdlib>
  4. #include <iostream>
  5. #include <algorithm>
  6. #include <queue>

  7. const int kMaxN = 20000 + 1;
  8. const int kMaxM = 100000 + 1;
  9. const int kInf = 0x7FFFFFFF;

  10. typedef struct graph *Graph;
  11. typedef struct node *Link;
  12. typedef struct edge Edge;
  13. struct graph {
  14.         int V;
  15.         int E;
  16.         Link *adj;
  17. };
  18. struct node {
  19.         int v;
  20.         int len;
  21.         Link next;
  22. };
  23. struct edge {
  24.         int v;
  25.         int w;
  26.         int len;
  27. };

  28. struct prison {
  29.         int v;
  30.         int w;
  31.         int len;
  32. } people[kMaxM];

  33. int n, m;

  34. inline Graph GraphBuild(int v) {
  35.         Graph t = (Graph)malloc(sizeof *t);

  36.         t->V = v;
  37.         t->E = 0;
  38.         t->adj = (Link *)malloc((v + 1) * sizeof(Link));
  39.         for (int i = 0; i <= v; i++) {
  40.                 t->adj[i] = NULL;
  41.         }

  42.         return t;
  43. }

  44. inline void GraphReset(Graph& G) {
  45.         G->E = 0;
  46.         for (int i = 0; i <= G->V; i++) {
  47.                 G->adj[i] = NULL;
  48.         }
  49. }

  50. inline Link NewEdge(int v, int len, Link next) {
  51.         Link t = (Link)malloc(sizeof *t);

  52.         t->v = v;
  53.         t->len = len;
  54.         t->next = next;

  55.         return t;
  56. }

  57. inline Edge EdgeNew(int v, int w, int len) {
  58.         Edge e;

  59.         e.v = v;
  60.         e.w = w;
  61.         e.len = len;

  62.         return e;
  63. }

  64. inline void GraphInsert(Graph &G, Edge e) {
  65.         int v = e.v;
  66.         int w = e.w;
  67.         int len = e.len;

  68.         G->adj[v] = NewEdge(w, len, G->adj[v]);
  69.         G->adj[w] = NewEdge(v, len, G->adj[w]);
  70.         G->E++;
  71. }

  72. inline bool BinaryGraphJudge(Graph &G) {
  73.         std::queue<int> q;
  74.         bool visited[kMaxN];
  75.         int vertex_color[kMaxN];
  76.         int visited_vertex = 0;

  77.         memset(visited, false, sizeof(visited));
  78.         memset(vertex_color, -1, sizeof(vertex_color));

  79.         q.push(1);
  80.         vertex_color[1] = 1;
  81.         while (visited_vertex < G->V) {
  82.                 if (q.empty()) {
  83.                         for (int i = 1; i <= G->V; i++) {
  84.                                 if (!visited[i]) {
  85.                                         q.push(i);
  86.                                         break;
  87.                                 }
  88.                         }
  89.                 }

  90.                 while (!q.empty()) {
  91.                         int now;
  92.                         Link t;

  93.                         now = q.front();
  94.                         q.pop();
  95.                         visited[now] = true;
  96.                         visited_vertex++;
  97.                         for (t = G->adj[now]; t != NULL; t = t->next) {
  98.                                 if (vertex_color[t->v] == -1 || vertex_color[t->v] == 3 - vertex_color[now]) {
  99.                                         vertex_color[t->v] = 3 - vertex_color[now];
  100.                                 } else {
  101.                                         return false;
  102.                                 }

  103.                                 if (!visited[t->v]) {
  104.                                         q.push(t->v);
  105.                                 }
  106.                         }
  107.                 }
  108.         }

  109.         return true;
  110. }

  111. inline void AddGraph(Graph &G, int mid_border) {
  112.         GraphReset(G);

  113.         for (int i = mid_border; i <= m; i++) {
  114.                 GraphInsert(G, EdgeNew(people[i].v, people[i].w, people[i].len));
  115.         }
  116. }

  117. inline int BinarySearch(Graph &G, int left_border, int right_border) {
  118.         int mid_border;

  119.         while (left_border < right_border) {
  120.                 mid_border = (left_border + right_border) / 2;
  121.                 
  122.                 AddGraph(G, mid_border);

  123.                 if (BinaryGraphJudge(G)) {
  124.                         right_border = mid_border;
  125.                 } else {
  126.                         left_border = mid_border + 1;
  127.                 }
  128.         }

  129.         return left_border - 1;
  130. }

  131. bool compare(const prison &a, const prison &b) {
  132.         return a.len < b.len;
  133. }

  134. int main()
  135. {
  136.         int v, w, len;
  137.         int ans;
  138.         Graph G;

  139.         std::ios::sync_with_stdio(false);
  140.         std::cin >> n;
  141.         std::cin >> m;
  142.         
  143.         G = GraphBuild(n);
  144.         
  145.         people[0].v = -1;
  146.         people[0].w = -1;
  147.         people[0].len = 0;
  148.         for (int i = 1; i <= m; i++) {
  149.                 std::cin >> v >> w >> n;
  150.                 people[i].v = v;
  151.                 people[i].w = w;
  152.                 people[i].len = len;
  153.         }

  154.         std::sort(people, people + 1 + m, compare);
  155.         
  156.         ans = BinarySearch(G, 0, m);
  157.         std::cout << people[ans].len << std::endl;

  158.         return 0;
  159. }
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