UVA 140 Brandwidth 带宽 (dfs回溯)
看到next_permutation好像也能过╮(╯▽╰)╭
这题学习点:
1.建图做映射
2.通过定序枚举保证字典序最小
3.strtok,sscanf,strchr等函数又复习了一遍,尽管程序中没有实际用上
4.剪枝,或者回溯
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<bits/stdc++.h> using namespace std; int G[8][8],deg[8]; bool vG[8][8];//判读连通 int pos[8]; bool vis[8]; int k; int best[8]; int cnt; int ID[26]; char rev_ID[8]; void dfs(int d,int width) { if(d == cnt){ if(width < k){ k = width; memcpy(best,pos,sizeof(pos)); } return; } for(int i = 0;i < cnt; i++) if(!vis[i]){//把i放在d位置 pos[d] = i; //prune 计算i和之前确定位置的结点的最大距离 int m = 0; for(int j = d-1; j >=0; j--) if(vG[i][pos[j]]) { m = max(m,d-j); if(m >= k) continue;//这题数据水了,这里写成return还是能过 } //prune2 计算u和未确定位置的相邻点的个数,适用于简单图 int ct = 0; for(int j = 0; j < deg[i]; j++) if(!vis[G[i][j]]) ct++; if(ct >= k) continue; vis[i] = 1; dfs(d+1,max(width,m)); vis[i] = 0; } } int main() { // freopen("in.txt","r",stdin); char buf[100]; const int INF = 0x3fffffff; while(fgets(buf,100,stdin) && *buf!='#'){ cnt = 0; memset(vG,0,sizeof(vG)); memset(deg,0,sizeof(deg)); memset(ID,-1,sizeof(ID)); memset(vis,0,sizeof(vis)); bool ap[26] ; memset(ap,0,sizeof(ap)); for(char *cur = buf; *cur ; cur++) { int u = *cur - 'A'; if(0<= u && u < 26 ) ap[u] = 1; } for(int i = 0; i < 26; i++){ if(ap[i]){ rev_ID[cnt] = i+'A'; ID[i] = cnt++; } } for(char *cur = buf; *cur ; cur++) { int u = ID[*cur - 'A']; for(cur+=2 ; *cur != '\n' && *cur != ';' ; cur++) { int v = ID[*cur - 'A']; vG[u][v] = vG[v][u] = 1; } } for(int i = 0; i < cnt; i++) for(int j = 0; j < cnt; j++){ if(vG[i][j]) G[i][deg[i]++] = j; } k = INF; dfs(0,0); for(int i = 0; i < cnt; i++){ printf("%c ",rev_ID[best[i]]); } printf("-> %d\n",k); } return 0; }