ZOJ 3332 Strange Country II (竞赛图构造哈密顿通路)

链接:http://www.icpc.moe/onlinejudge/showProblem.do?problemCode=3332

本文链接:http://www.cnblogs.com/Ash-ly/p/5454611.html

题意:

　　给你一个N,代表含有N个点的竞赛图,接着的N * (N- 1) / 2行两个点u, v,代表存在有向边<u,v>,问是否能构造出来一个哈密顿通路.

思路:

　　竞赛图一定含有哈密顿通路,不一定含有哈密顿回路.则不可能出现不存在的情况,直接构造就可以,至于方法可看我的另外一篇文章:http://www.cnblogs.com/Ash-ly/p/5452580.html.

注意:

　　构造的时候从后往前寻找1或者0的时候一定是按照当前序列的顺序查找到,而不是按照点本身的顺序查找,在这里WA了几次.

代码:

 #include <iostream>
 #include <cmath>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>
 #include <queue>
 #include <stack>
 #include <vector>

 using namespace std;
 typedef long long LL;
 const int maxN = ;

 //The arv[] length is len, insert key befor arv[index]
 inline void Insert(int arv[], int &len, int index, int key){
     if(index > len) index = len;
     len++;
     for(int i = len - ; i >= ; --i){
         if(i != index && i)arv[i] = arv[i - ];
         else{arv[i] = key; return;}
     }
 }

 void Hamilton(int ans[maxN + ], int map[maxN + ][maxN + ], int n){
     int ansi = ;
     ans[ansi++] = ;
     for(int i = ; i <= n; i++){
         if(map[i][ans[ansi - ]] == )
             ans[ansi++] = i;
         else{
             int flag = ;
             for(int j = ansi - ; j > ; --j){//在当前序列中查找0/1
                 if(map[i][ans[j]] == ){
                     flag = ;
                     Insert(ans, ansi, j + , i);
                     break;
                 }
             }
             if(!flag)Insert(ans, ansi, , i);
         }
     }
 }

 int main()
 {
     //freopen("input.txt", "r", stdin);
     int t;
     scanf("%d", &t);
     while(t--){
         int N;
         scanf("%d", &N);
         int M = N * (N - ) / ;
         int map[maxN + ][maxN + ] = {};
         for(int i = ; i < M; i++){
             int u, v;
             scanf("%d%d", &u, &v);
             if(u < v)map[v][u] = ;//建图,map[v][u] = 1,代表存在边<u, v>,且 u < v.
         }
         int ans[maxN + ] = {};
         Hamilton(ans, map, N);
         for(int i = ; i <= N; i++)
             printf(i ==  ? "%d":" %d", ans[i]);
         printf("\n");
     }
     return ;
 }

代码2:

 #include <iostream>
 #include <cmath>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>
 #include <queue>
 #include <stack>
 #include <vector>

 using namespace std;
 typedef long long LL;
 const int maxN = ;

 inline void read(int&a){char c;while(!(((c=getchar())>='')&&(c<='')));a=c-'';while(((c=getchar())>='')&&(c<=''))(a*=)+=c-'';}

 void Hamilton(int ans[maxN + ], int map[maxN + ][maxN + ], int n){
     int nxt[maxN + ];//用数组模拟链表
     memset(nxt, -, sizeof(nxt));
     int head = ;
     for(int i = ; i <= n; i++){
         if(map[i][head]){
             nxt[i] = head;
             head = i;
         }else{
             int pre = head, pos = nxt[head];
             while(pos != - && !map[i][pos]){
                 pre = pos;
                 pos = nxt[pre];
             }
             nxt[pre] = i;
             nxt[i] = pos;
         }
     }
     int cnt = ;
     for(int i = head; i != -; i = nxt[i])
         ans[++cnt] = i;
 }

 int main()
 {
     freopen("input.txt", "r", stdin);
     int N;
     while(~scanf("%d", &N)){
         int map[maxN + ][maxN + ] = {};
         for(int i = ; i <= N; i++){
             for(int j = ; j <= N; j++){
                 int u;
                 read(u);
                 map[i][j] = u;
             }
         }
         printf("%d\n%d\n", , N);
         int ans[maxN + ] = {};
         Hamilton(ans, map, N);
         for(int i = ; i <= N; i++)
             printf(i ==  ? "%d":" %d", ans[i]);
         printf("\n");
     }
     return ;
 }