UVA140-Bandwidth（搜索剪枝）

Posted 2021-01-01 npugen

Problem UVA140-Bandwidth

Time Limit: 3000 mSec

 Problem Description

 

Given a graph (V, E) where V is a set of nodes and E is a set of arcs in V ×V , and an ordering on the elements in V , then the bandwidth of a node v is defined as the maximum distance in the ordering between v and any node to which it is connected in the graph. The bandwidth of the ordering is then defined as the maximum of the individual bandwidths. For example, consider the graph on the right: This can be ordered in many ways, two of which are illustrated below: For these orderings, the bandwidths of the nodes (in order) are 6, 6, 1, 4, 1, 1, 6, 6 giving an ordering bandwidth of 6, and 5, 3, 1, 4, 3, 5, 1, 4 giving an ordering bandwidth of 5. Write a program that will find the ordering of a graph that minimises the bandwidth.

 

 Input

Input will consist of a series of graphs. Each graph will appear on a line by itself. The entire file will be terminated by a line consisting of a single ‘#’. For each graph, the input will consist of a series of records separated by ‘;’. Each record will consist of a node name (a single upper case character in the the range ‘A’ to ‘Z’), followed by a ‘:’ and at least one of its neighbours. The graph will contain no more than 8 nodes.

 

 Output

Output will consist of one line for each graph, listing the ordering of the nodes followed by an arrow (->) and the bandwidth for that ordering. All items must be separated from their neighbours by exactly one space. If more than one ordering produces the same bandwidth, then choose the smallest in lexicographic ordering, that is the one that would appear first in an alphabetic listing.

 

 Sample Input

A:FB;B:GC;D:GC;F:AGH;E:HD

#

 

 Sample Ouput

A B C F G D H E -> 3

 

题解：回溯法，剪枝主要有两点：最优化剪枝，这是显然的；如果搜索到u节点，此时u节点的孩子节点还有m个没有确定，那么最理想的情况下带宽也至少是m，因此如果m > 当前最小带宽，就完全可以剪枝。

 

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cstdlib>
  5 #include <vector>
  6 #define INF 0x3f3f3f3f
  7 using namespace std;
  8 
  9 const int maxl = 200+10;
 10 const int kind = 26;
 11 char str[maxl];
 12 int n,Min = INF,id[256];
 13 int res[maxl];
 14 char converse[kind];
 15 bool vis[kind];
 16 vector< vector<int> > child(kind);
 17 
 18 int pos_find(const int *num,int len,int tar){
 19     for(int k = 0;k < len;k++){
 20         if(num[k] == tar) return k;
 21     }
 22     return -1;
 23 }
 24 
 25 bool check(const int *ans,int &wide){
 26     int u,v;
 27     for(int i = 0;i < n;i++){
 28         u = ans[i];
 29         for(int j = 0;j < child[u].size();j++){
 30             v = child[u][j];
 31             int pos = pos_find(ans,n,v);
 32             wide = max(wide,abs(i-pos));
 33             if(wide > Min) return false;
 34         }
 35     }
 36     return true;
 37 }
 38 
 39 void dfs(int *ans,int cur,int wide){
 40     //printf("cur:%d
",cur);
 41     if(wide > Min) return;
 42     if(cur == n){
 43         if(check(ans,wide)){
 44             if(wide < Min){
 45                 Min = wide;
 46                 memcpy(res,ans,n*sizeof(int));
 47             }
 48             else if(wide == Min){
 49                 if(res[0] == -1) memcpy(res,ans,n*sizeof(int));
 50                 else{
 51                     int p = 0;
 52                     while(res[p] == ans[p]) p++;
 53                     if(p!=n && ans[p]<res[p]) memcpy(res,ans,n*sizeof(int));
 54                 }
 55             }
 56         }
 57         return;
 58     }
 59     for(int u = 0;u < n;u++){
 60         if(!vis[u]){
 61             int cnt = 0,tmp = wide;
 62             bool ok = true;
 63             for(int j = 0;j < child[u].size();j++){
 64                 int v = child[u][j];
 65                 if(!vis[v]) continue;
 66                 cnt++;
 67                 int pos = pos_find(ans,cur,v);
 68                 tmp = max(wide,cur-pos);
 69                 //printf("tmp:%d
",tmp);
 70                 if(tmp > Min){
 71                     ok = false;
 72                     break;
 73                 }
 74             }
 75             if(!ok) continue;
 76             cnt = child[u].size()-cnt;
 77             if(cnt > Min) continue;
 78             ans[cur] = u;
 79             vis[u] = true;
 80             dfs(ans,cur+1,tmp);
 81             vis[u] = false;
 82         }
 83     }
 84 }
 85 
 86 int main()
 87 {
 88     //freopen("input.txt","r",stdin);
 89     //freopen("output.txt","w",stdout);
 90     while(~scanf("%s",str) && str[0]!=‘#‘){
 91         n = 0;
 92         Min = INF;
 93         memset(res,-1,sizeof(res));
 94         memset(vis,false,sizeof(vis));
 95         int len = strlen(str);
 96         for(char ch = ‘A‘;ch <= ‘Z‘;ch++){
 97             if(strchr(str,ch) != NULL){
 98                 id[ch-‘‘] = n++;
 99                 converse[n-1] = ch;
100             }
101         }
102         for(int i = 0;i < n;i++) child[i].clear();
103         for(int i = 0;i < len;i++){
104             int u = id[str[i]-‘‘];
105             i += 2;
106             while(str[i]!=‘;‘ && i<len){
107                 int v = id[str[i++]-‘‘];
108                 child[u].push_back(v);
109                 child[v].push_back(u);
110             }
111         }
112         //for(int i = 0;i < n;i++) printf("%d
",child[i].size());
113         int ans[kind];
114         memset(ans,0,sizeof(ans));
115         dfs(ans,0,0);
116         for(int i = 0;i < n;i++){
117             printf("%c ",converse[res[i]]);
118         }
119         printf("-> %d
",Min);
120     }
121     return 0;
122 }