SGU101 DFS

    题意：给你一些Domino，两端有数字，问能否用全部骨牌首尾相连成一条链，使得相连的两端数字相同。可以的话要给出方案。

    Problem：Give you some dominoes which have numbers on it's both sides. Can you connect the dominoes one by one to form a chain? It requires the dominoes touch through equal marked sides. If there exist any solution, you should give one.

    解法：先算出现次数，显然出现奇数次的牌必须是0个或2个。之后DFS，使得每个牌都用上。最后再根据DFS顺序得到方案。

    Solution：First calculate the times that these numbers appeared. Apparently, the number of dominoes which appear odd times must be 0 or 2.  Then DFS to make every domino to be used. Finally, you make the solution according to the DFS result.

    

#include <iostream>#include <stdio.h>#include <cstring>using namespace std;struct data {       int x,y;}g[1000],ans[1000];int x[10],a[10][10],n,cnt,s,b,c;void dfs(int x) {     for (int i=0;i<=6;i++)         if (a[x][i]) {            a[x][i]--;            a[i][x]--;            dfs(i);            cnt++;            ans[cnt].x = x;            ans[cnt].y = i;         }}int main() {    while (~scanf("%d",&n)) {          memset(a,0,sizeof(a));          memset(x,0,sizeof(x));          for (int i=1;i<=n;i++) {              scanf("%d%d",&b,&c);              a[b][c]++;              a[c][b]++;              x[b]++;              x[c]++;              g[i].x = b;              g[i].y = c;          }          cnt = 0, s = -1;          for (int i=0;i<=6;i++) {              if (x[i]%2==1) {                 cnt++;                 s = i;              }              else if (x[i] && s==-1) s = i;          }          if (cnt==1 || cnt>2) {             printf("No solution\n");             continue;          }          cnt = 0;          dfs(s);          if (cnt<n) {             printf("No solution\n");             continue;          }          for (int i=cnt;i>=1;i--)              for (int j=1;j<=n;j++)                  if (ans[i].x==g[j].x && ans[i].y==g[j].y) {                     printf("%d +\n",j);                     g[j].x = -1;                     break;                  }                  else if (ans[i].x==g[j].y && ans[i].y==g[j].x) {                     printf("%d -\n",j);                     g[j].x = -1;                     break;                  }    }    return 0;}