hdu 3551 Hard Problem 一般圖最大匹配+給出一個無向圖，存在重邊，沒有自環。問能否刪除一些邊，使得每個頂點的度數爲指定度數

Problem Description

This is the most tough task in this contest, do not try it until you solve all the other tasks or you feel boring on others. Given an undirected graph, you are to find out a subgraph of it so that the degree of the i-th node in the subgraph is the given integer D_i. The subgraph is a subset of edges and all vertexes are reserved. Notice that the graph may be disconnected, and two edges may connect the same vertexes, but no self cyclic exists.

 

 

Input

The input contains several test cases, the first line of the input contains an integer T denoting the number of test cases.
For each test case, the first line contains two integers N and M, denoting the number of vertexes and number of edges in the graph. (1 <= N <= 50, 1 <= M <= 200)
For the next M lines, each line contains two integers X and Y, denote there is a edge between X-th node and Y-th node. (1 <= X, Y <= N)
For the last N lines, each line contains a single integer D_i, denote the degree of i-th node in the subgraph.

 

 

Output

For each test case, if the subgraph exist, output "YES", otherwise output "NO". See sample output for further details

 

 

Sample Input

2 4 4 1 2 3 4 2 3 1 4 1 2 1 0 4 5 2 1 1 2 2 3 3 4 3 4 1 0 1 0

 

 

Sample Output

Case 1: YES Case 2: NO

 

 //

 我們應該刪除哪些邊呢？ 預處理每個頂點的度數d[i], 若d[i] = deg[i], 那麼 與這個點相連的邊是不能刪掉的。原因很顯然。若i與j之間有邊，並且d[i]>deg[i], d[j]>deg[j]那麼這條邊是可以刪除的。接下來如何建圖呢？ 將每個點i 拆成 d[i] - deg[i]個點。如果i與j之間的邊e可以刪除, 則邊e與i、j拆出的每個點連一條邊 ei, ej（重邊連多次）。然後求該一般圖最大匹配，若存在完美匹配，則YES。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int MAXN = 800;
int n,head,tail,start,finish,match[MAXN],fa[MAXN],base[MAXN],Q[MAXN];//match初始爲0 n個點的圖 從1開始
int adj[MAXN][MAXN];   //鄰接矩陣
bool mark[MAXN],in_blossom[MAXN],in_queue[MAXN];
inline void Contract (int x,int y){
  memset(mark,0,sizeof(mark));
  memset(in_blossom,0,sizeof(in_blossom));
#define pre fa[match[i]]
  int lca,i;
  for(i = x;i;i = pre){
    i = base[i];
    mark[i] = 1;
  }
  for(i = y;i; i = pre){
    i = base[i];
    if(mark[i]){
      lca = i;
      break;
    }
  }
  for (i = x; base[i] != lca; i = pre){
    if(base[pre] != lca) fa[pre] = match[i];
    in_blossom[base[i]] = 1;
    in_blossom[base[match[i]]] = 1;
  }
  for (i = y; base[i] != lca; i = pre){
    if (base[pre] != lca) fa[pre] = match[i];
    in_blossom[base[i]] = 1;
    in_blossom[base[match[i]]] = 1;
  }
#undef pre
  if (base[x] != lca) fa[x] = y;
  if (base[y] != lca) fa[y] = x;
  for (i = 1; i <= n; ++i){
    if (in_blossom[base[i]]){
      base[i] = lca;
      if (!in_queue[i]){
        Q[++tail] = i;
        in_queue[i] = 1;
      }
    }
  }
}
inline void Change(){
  int x,y,z;
  z = finish;
  while (z){
    y = fa[z];
    x = match[y];
    match[y] = z;
    match[z] = y;
    z = x;
  }
}
inline void FindAugmentPath(){
  memset(fa,0,sizeof(fa));
  memset(in_queue,0,sizeof(in_queue));
  for(int i = 1; i <= n; ++i)base[i] = i;
  head = 0; tail = 1;
  Q[1] = start;
  in_queue[start] = 1;
  while (head != tail){
    int x = Q[++head];
    for (int y = 1; y <= n; ++y){
      if (adj[x][y] && base[x] != base[y] && match[x] != y)
        if (start == y || match[y] && fa[match[y]])
          Contract(x,y);
        else if(!fa[y]){
          fa[y] = x;
          if(match[y]){
            Q[++tail] = match[y];
            in_queue[match[y]] = 1;
          }
          else {
            finish = y;
            Change();
            return;
          }
        }
    }
  }
}
inline void Edmonds(){
  memset(match,0,sizeof(match));
  for (start = 1; start <= n; ++start)
    if (match[start] == 0)
      FindAugmentPath();
}
inline void init(){
 memset(adj,0,sizeof(adj));
}
int deg[MAXN], D[MAXN], M;
pair<int ,int> edge[MAXN], id[MAXN];
int main(){
  int Case, u, v,V;
  scanf("%d",&Case);
  for(int it = 1;it <= Case; ++it){
    scanf("%d%d",&V,&M);
    memset(deg, 0, sizeof(deg));
    memset(adj, 0, sizeof(adj));
    for(int i = 0;i < M; ++i){
      scanf("%d%d",&u,&v);
      u --; v --;
      edge[i] = make_pair(u, v);
      deg[u] ++; deg[v] ++;
    }
    for(int i = 0;i < V; ++i){
      scanf("%d",&D[i]);
    }
    bool flag = true;
    int cnt = 1;
    for(int i = 0;i < MAXN; ++i) id[i] = make_pair(-1, -1);
    for(int i = 0;i < V; ++i)
      if(deg[i] < D[i]) { flag = false; break; }
    printf("Case %d: ",it);
    if(!flag) {
      puts("NO");
      continue;
    }
    for(int i = 0;i < M; ++i){
      u = edge[i].first;
      v = edge[i].second;
      if(id[u].first == -1){
        id[u] = make_pair(cnt, cnt + deg[u] - D[u] - 1);
        cnt += (deg[u] - D[u]);
      }
      if(id[v].first == -1){
        id[v] = make_pair(cnt, cnt + deg[v] - D[v] - 1);
        cnt += (deg[v] - D[v]);
      }
      if(id[V+i].first == -1){
        id[V+i] = make_pair(cnt, cnt + 1);
        cnt += 2;
      }
      int t = id[V+i].first;
      adj[t][t+1] = adj[t+1][t] = true;//important
      for(int j = id[u].first;j <= id[u].second; ++j)
        adj[t][j] = adj[j][t] = true;
      for(int j = id[v].first;j <= id[v].second; ++j)
        adj[t+1][j] = adj[j][t+1] = true;
    }
    int j, sum = 0;
    n  = cnt-1;
    flag = 1;
    Edmonds();
    for(int i = 1; i <= n; ++i){
      if(!match[i]){
        flag = 0;
        break;
      }
    }
    if(flag) puts("YES");
    else puts("NO");
  }
}