每個字符串至少出現兩次且不重疊的最長子串
阿新 • • 發佈：2018-08-18
href div title eight mes 。。 scan truct oid
Relevant Phrases of Annihilation
SPOJ - PHRASES
https://cn.vjudge.net/problem/SPOJ-PHRASES
呵。。。呵。。。
我覺得我寫的很對呀。。。
真是的。。。
這麽漂亮。。。
行吧。。一下午沒檢查出來哪裏有問題 真是的 太過分了。。。emm。。。
過幾天再看。。啊啊啊。。。
網上正確代碼：
#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include<cctype> #include <set> #include <vector> #include <stack> #include <queue> #include <algorithm> #include <cmath> #define rap(i, a, n) for(int i=a; i<=n; i++) #define rep(i, a, n) for(int i=a; i<n; i++) #define lap(i, a, n) for(int i=n; i>=a; i--) #definelep(i, a, n) for(int i=n; i>a; i--) #define rd(a) scanf("%d", &a) #define rlld(a) scanf("%lld", &a) #define rc(a) scanf("%c", &a) #define rs(a) scanf("%s", a) #define MOD 2018 #define LL long long #define ULL unsigned long long #define Pair pair<int, int> #define mem(a, b) memset(a, b, sizeof(a)) #define_ ios_base::sync_with_stdio(0),cin.tie(0) //freopen("1.txt", "r", stdin); using namespace std; const int maxn = 101000, INF = 0x7fffffff; int s[maxn]; int sa[maxn], t[maxn], t2[maxn], c[maxn], n; int ran[maxn], height[maxn], length[105]; void get_sa(int m) { int i, *x = t, *y = t2; for(i = 0; i < m; i++) c[i] = 0; for(i = 0; i < n; i++) c[x[i] = s[i]]++; for(i = 1; i < m; i++) c[i] += c[i-1]; for(i = n-1; i >= 0; i--) sa[--c[x[i]]] = i; for(int k = 1; k <= n; k <<= 1) { int p = 0; for(i = n-k; i < n; i++) y[p++] = i; for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k; for(i = 0; i < m; i++) c[i] = 0; for(i = 0; i < n; i++) c[x[y[i]]]++; for(i = 0; i< m; i++) c[i] += c[i-1]; for(i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i]; swap(x, y); p = 1; x[sa[0]] = 0; for(i = 1; i < n; i++) x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k] ? p-1 : p++; if(p >= n) break; m = p; } int k = 0; for(i = 0; i < n; i++) ran[sa[i]] = i; for(i = 0; i < n; i++) { if(k) k--; int j = sa[ran[i]-1]; while(s[i+k] == s[j+k]) k++; height[ran[i]] = k; } } char str[maxn]; int id[maxn]; struct node { int Min; int Max; int flag; } anspos[15]; void ini() { for(int i = 0; i <= 12; i++) anspos[i].Min = 0x3f3f3f3f,anspos[i].Max = -1; } bool can(int len,int n,int num) { int l = 2,r = 2; ini(); for(int i = 2; i <= n; i++) { if(height[i] >= len) { int id1=id[sa[i-1]]; int id2=id[sa[i]]; anspos[id1].Max=max(anspos[id1].Max,sa[i-1]); anspos[id1].Min=min(anspos[id1].Min,sa[i-1]); anspos[id2].Max=max(anspos[id2].Max,sa[i]); anspos[id2].Min=min(anspos[id2].Min,sa[i]); int t; for(t = 0; t < num; t++) if(anspos[t].Max - anspos[t].Min < len) break; if(t == num) return 1; } else { for(int j = 0; j <= 10; j++) anspos[j].Min = 0x3f3f3f3f,anspos[j].Max = -1; } } for(int i = 0; i < num; i++) if(!anspos[i].flag) return 0; return 1; } int main() { int k,q,t; scanf("%d",&t); while(t--) { ini(); int tot = 0,len = 0x3f3f3f3f; scanf("%d",&q); for(int i = 0; i < q; i++) { scanf("%s",str); for(int j = 0; str[j]!=‘\0‘; j++) { id[tot] = i; s[tot++] = str[j]; } id[tot] = i; s[tot++] = ‘#‘+i; len = min(len,(int)strlen(str)); } s[tot] = 0; n = tot+1; get_sa(200); // for(int i=1; i<n; i++) // cout<< height[i] <<endl; int l = 0,r = len; int ans = 0; while(l <= r) { int mid = l + (r - l) / 2; if(can(mid,tot,q)) { l = mid+1; } else r = mid-1; } printf("%d\n",r); } return 0; }
我的：
#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <cctype> #include <set> #include <vector> #include <stack> #include <queue> #include <algorithm> #include <cmath> #define rap(i, a, n) for(int i=a; i<=n; i++) #define rep(i, a, n) for(int i=a; i<n; i++) #define lap(i, a, n) for(int i=n; i>=a; i--) #define lep(i, a, n) for(int i=n; i>a; i--) #define rd(a) scanf("%d", &a) #define rlld(a) scanf("%lld", &a) #define rc(a) scanf("%c", &a) #define rs(a) scanf("%s", a) #define MOD 2018 #define LL long long #define ULL unsigned long long #define Pair pair<int, int> #define mem(a, b) memset(a, b, sizeof(a)) #define _ ios_base::sync_with_stdio(0),cin.tie(0) //freopen("1.txt", "r", stdin); using namespace std; const int maxn = 101000, INF = 0x7fffffff; int s[maxn]; int sa[maxn], t[maxn], t2[maxn], c[maxn], n; int ran[maxn], height[maxn], length[105]; void get_sa(int m) { int i, *x = t, *y = t2; for(i = 0; i < m; i++) c[i] = 0; for(i = 0; i < n; i++) c[x[i] = s[i]]++; for(i = 1; i < m; i++) c[i] += c[i-1]; for(i = n-1; i >= 0; i--) sa[--c[x[i]]] = i; for(int k = 1; k <= n; k <<= 1) { int p = 0; for(i = n-k; i < n; i++) y[p++] = i; for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k; for(i = 0; i < m; i++) c[i] = 0; for(i = 0; i < n; i++) c[x[y[i]]]++; for(i = 0; i< m; i++) c[i] += c[i-1]; for(i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i]; swap(x, y); p = 1; x[sa[0]] = 0; for(i = 1; i < n; i++) x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k] ? p-1 : p++; if(p >= n) break; m = p; } int k = 0; for(i = 0; i < n; i++) ran[sa[i]] = i; for(i = 0; i < n; i++) { if(k) k--; int j = sa[ran[i]-1]; while(s[i+k] == s[j+k]) k++; height[ran[i]] = k; } } struct node { int minn, maxx; }Node[15]; int solve(int k, int q) { for(int i=1; i<=q; i++) Node[i].minn = INF, Node[i].maxx = -INF; int cnt = 0; for(int i=1; i<n; i++) { if(height[i] < k) { for(int j=1; j<=q; j++) { Node[j].minn = INF, Node[j].maxx = -INF; } continue; } for(int j=1; j<=q; j++) { if(sa[i] > length[j-1] && sa[i] < length[j]) { Node[j].minn = min(Node[j].minn, sa[i]); Node[j].maxx = max(Node[j].maxx, sa[i]); } if(sa[i-1] > length[j-1] && sa[i-1] < length[j]) { Node[j].minn = min(Node[j].minn, sa[i-1]); Node[j].maxx = max(Node[j].maxx, sa[i-1]); } } cnt = 0; for(int j=1; j<=q; j++) { if(Node[j].maxx - Node[j].minn >= k) cnt++; } if(cnt == q) return true; } return false; } int T, q, len; char str[maxn]; int main() { rd(T); while(T--) { n = 0; int l = 0, r = INF; length[0] = 0; rd(q); rep(i, 0, q) { rs(str); len = strlen(str); r = min(r, len); rep(j, 0, len) s[n++] = str[j]; length[i+1] = n; s[n++] = ‘#‘ + i; } s[n++] = 0; get_sa(200); // for(int i=1; i<n; i++) // cout<< height[i] <<endl; // cout<< l << " " << r <<endl; while(l <= r) { int mid = l + (r - l) / 2; if(solve(mid, q)) l = mid + 1; else r = mid - 1; // cout<< 111 <<endl; } cout<< r <<endl; } return 0; }
每個字符串至少出現兩次且不重疊的最長子串