标签：str back ORC cdb pac fopen string == turn
题解：
枚举非法对。
如果 ‘a‘ 和 ’b‘ 不能相邻的话，那么删除 ‘a‘ ‘b‘之间的字符就是非法操作了。 假设题目给定的字符串为 "acdbe"，所以删除cd是非法操作， 因为cd是非法了，所以cde也是非法操作，
也就是说找到所有的非法操作之后往外推，比他多删的状态就一样是非法的了，当然对于上述的“acdbe"来说，不能确定 ”acd"是非法操作，因为在枚举非法对的时候，该非法对的字符并不能被删除。
然后把所有非法对的非法状态都存下来。然后从没删除的状态往外搜，就好了。
代码：
/* code by: zstu wxk time: 2019/03/06 Problem Link: http://codeforces.com/contest/1117/problem/F Solve: */ #include<bits/stdc++.h> using namespace std; #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); #define LL long long #define ULL unsigned LL #define fi first #define se second #define pb push_back #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define lch(x) tr[x].son[0] #define rch(x) tr[x].son[1] #define max3(a,b,c) max(a,max(b,c)) #define min3(a,b,c) min(a,min(b,c)) typedef pair<int,int> pll; const int inf = 0x3f3f3f3f; const int _inf = 0xc0c0c0c0; const LL INF = 0x3f3f3f3f3f3f3f3f; const LL _INF = 0xc0c0c0c0c0c0c0c0; const LL mod = (int)1e9+7; const int N = 1e5 + 100; char s[N]; int a[20][20], cnt[20]; int tmp[1<<18], gg[1<<18]; int n, m, k; void solve(int u, int v){ memset(tmp, 0, sizeof(tmp)); for(int i = 1, j; i <= n;){ while(i <= n && s[i]-‘a‘ != u) ++i; int t = 0; j = i+1; while(j <= n && s[j] -‘a‘ != u && s[j]-‘a‘ != v) { t |= 1 << s[j]-‘a‘; ++j; } if(j > n) break; if(s[j] - ‘a‘ == v){ i = j+1; tmp[t] = 1; } if(s[j] - ‘a‘ == u) i = j; } for(int i = 0; i <= k; ++i){ if(tmp[i]){ gg[i] = 1; for(int j = 0; j < m; ++j){ if(j == u || j == v) continue; tmp[i|(1<<j)] = 1; } } } return ; } int vis[1<<18]; int ans; void DFS(int x, int len){ if(vis[x]) return ; vis[x] = 1; ans = min(ans, len); for(int i = 0; i < m; ++i){ if(gg[x | (1<<i)]) continue; DFS(x|1<<i, len-cnt[i]); } return ; } void Ac(){ memset(gg, 0, sizeof(gg)); memset(cnt, 0, sizeof cnt); scanf("%s", s+1); for(int i = 0; i < m; ++i) for(int j = 0; j < m; ++j) scanf("%d", &a[i][j]); k = (1<<m) - 1; for(int i = 0; i < m; ++i) for(int j = 0; j < m; ++j) if(a[i][j] == 0) solve(i,j); for(int i = 1; i <= n; ++i){ cnt[s[i]-‘a‘]++; } ans = n; DFS(0, n); printf("%d\n", ans); return ; } int main(){ while(~scanf("%d%d", &n, &m)){ Ac(); } return 0; }
标签：str back ORC cdb pac fopen string == turn
原文地址：https://www.cnblogs.com/MingSD/p/10496334.html
