题意：一个句子中有多个单词，但是目前的单词，除了首末两位，中间的单词字母顺序均被打乱，并且打乱后把单词间的空格删掉变成一个新句子。现在给定这个新句子（长度为1~1000），给定n个单词（1 <= n <= 10000），且每个单词是唯一的，求是否能用这n个单词还原出这个句子的原来的样子，若不能则输出"impossible"，若多解则输出"ambiguous"，否则输出这个句子。（例：tihssnetnceemkaesprfecetsesne，给定makes，perfect，sense，sentence，this，只有一种解读方式：this sentence makes perfect sense）
1、d[i]表示从开始到第 i 位这一段有几种构成方法；
2、从前往后找，看看对于某一位能不能往前延伸构成一个单词，设 i ~ j 位可以构成一个单词，则 d[j] += d[i - 1]，最后看d[len - 1]的情况即可；
3、对于每个单词记录前缀和后缀，以及，记录每个单词的字母个数和目标串截止到每一位的字母个数，方便减小枚举量；
4、边dp边记录达到状态转移的要求的某一位的来源。
总共两个AC代码，第一次900ms多差点超时，换种写法写之后只有62ms，还是能用char数组就用char数组表示字符串，string太浪费时间。
代码如下：
62ms
#include<cstdio>
#include<cstring>
#include<cctype>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<list>
#define fin freopen("in.txt", "r", stdin)
#define fout freopen("out.txt", "w", stdout)
#define pr(x) cout << #x << " : " << x << " "
#define prln(x) cout << #x << " : " << x << endl
#define Min(a, b) a < b ? a : b
#define Max(a, b) a < b ? b : a
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const double pi = acos(-1.0);
const double EPS = 1e-6;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const ll MOD = 1e9 + 7;
using namespace std;
#define NDEBUG
#include<cassert>
const int MAXN = 1000 + 10;
const int MAXT = 10000 + 10;
char s[MAXT][110];
struct Node{
int id, len, num[30];
char head;
Node(int ids, int le) : id(ids), len(le){
memset(num, 0, sizeof num);
for(int i = 0; i < len; ++i)
++num[s[id][i] - ‘a‘];
head = s[id][0];
}
};
int T, n, plen, d[MAXN], last[MAXN], thisidx[MAXN], thisidy[MAXN], o[MAXN][30];
char p[MAXN];
vector<Node> vec[30];
inline bool judge(Node &u, int l, int r){
if(l == 0){
for(int i = 0; i < 26; ++i)
if(u.num[i] != o[r][i]) return false;
return true;
}
else{
for(int i = 0; i < 26; ++i)
if(u.num[i] != o[r][i] - o[l - 1][i]) return false;
return true;
}
}
void init(){
for(int i = 0; i < 26; ++i) vec[i].clear();
memset(d, 0, sizeof d);
memset(thisidy, 0, sizeof thisidx);
memset(thisidx, 0, sizeof thisidy);
memset(last, -1, sizeof last);
memset(o[0], 0, sizeof o[0]);
++o[0][p[0] - ‘a‘];
for(int i = 1; i < plen; ++i){
for(int j = 0; j < 26; ++j) o[i][j] = o[i - 1][j];
++o[i][p[i] - ‘a‘];
}
}
int main(){
scanf("%d", &T);
while(T--){
scanf("%s", p);
plen = strlen(p);
init();
scanf("%d", &n);
for(int i = 0; i < n; ++i){
scanf("%s", s[i]);
int le = strlen(s[i]);
vec[s[i][le - 1] - ‘a‘].push_back(Node(i, le));
}
for(int i = 0; i < plen; ++i){
int lur = p[i] - ‘a‘;
for(int j = 0; j < vec[lur].size(); ++j){
Node &u = vec[lur][j];
int tmp = i - u.len + 1;
if(tmp == 0 && u.head == p[tmp] && judge(u, tmp, i)){
++d[i];
thisidx[i] = lur;
thisidy[i] = j;
}
else if(tmp > 0 && u.head == p[tmp] && judge(u, tmp, i) && d[tmp - 1]){
d[i] += d[tmp - 1];
last[i] = tmp - 1;
thisidx[i] = lur;
thisidy[i] = j;
}
}
}
if(!d[plen - 1]) printf("impossible\n");
else if(d[plen - 1] > 1) printf("ambiguous\n");
else{
bool flag = false;
stack<pair<int, int> > st;
for(int i = plen - 1; i != -1; i = last[i])
st.push(pair<int, int>(thisidx[i], thisidy[i]));
while(!st.empty()){
if(flag) printf(" ");
pair<int, int> tmp = st.top(); st.pop();
printf("%s", s[vec[tmp.first][tmp.second].id]);
flag = true;
}
printf("\n");
}
}
return 0;
}
967ms
#include<cstdio>
#include<cstring>
#include<cctype>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<list>
#define fin freopen("in.txt", "r", stdin)
#define fout freopen("out.txt", "w", stdout)
#define pr(x) cout << #x << " : " << x << " "
#define prln(x) cout << #x << " : " << x << endl
#define Min(a, b) a < b ? a : b
#define Max(a, b) a < b ? b : a
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const double pi = acos(-1.0);
const double EPS = 1e-6;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const ll MOD = 1e9 + 7;
using namespace std;
#define NDEBUG
#include<cassert>
const int MAXN = 1000 + 10;
const int MAXT = 10000 + 10;
struct Node{
string aftersort;
int id, len;
char head;
Node(string &s, int ids){
id = ids;
len = s.length();
aftersort = s;
if(len > 3) sort(aftersort.begin() + 1, aftersort.end() - 1);
head = aftersort[0];
}
};
int T, n, plen, d[MAXN], last[MAXN], thisidx[MAXN], thisidy[MAXN];
string s[MAXT], p;
vector<Node> vec[30];
inline bool judge(Node &u, int i){
string tmp = p.substr(i, u.len);
if(u.len >= 3) sort(tmp.begin() + 1, tmp.end() - 1);
return tmp == u.aftersort;
}
void init(){
for(int i = 0; i < 26; ++i) vec[i].clear();
memset(d, 0, sizeof d);
memset(thisidy, 0, sizeof thisidx);
memset(thisidx, 0, sizeof thisidy);
memset(last, -1, sizeof last);
}
int main(){
scanf("%d", &T);
while(T--){
init();
cin >> p;
plen = p.length();
scanf("%d", &n);
for(int i = 0; i < n; ++i){
cin >> s[i];
vec[*s[i].rbegin() - ‘a‘].push_back(Node(s[i], i));
}
for(int i = 0; i < plen; ++i){
int lur = p[i] - ‘a‘;
for(int j = 0; j < vec[lur].size(); ++j){
Node &u = vec[lur][j];
int tmp = i - u.len + 1;
if(tmp == 0 && u.head == p[tmp] && judge(u, tmp)){
++d[i];
thisidx[i] = lur;
thisidy[i] = j;
}
else if(tmp > 0 && u.head == p[tmp] && judge(u, tmp) && d[tmp - 1]){
d[i] += d[tmp - 1];
last[i] = tmp - 1;
thisidx[i] = lur;
thisidy[i] = j;
}
}
}
if(!d[plen - 1]) printf("impossible\n");
else if(d[plen - 1] > 1) printf("ambiguous\n");
else{
bool flag = false;
stack<pair<int, int> > st;
for(int i = plen - 1; i != -1; i = last[i])
st.push(pair<int, int>(thisidx[i], thisidy[i]));
while(!st.empty()){
if(flag) printf(" ");
pair<int, int> tmp = st.top(); st.pop();
printf("%s", s[vec[tmp.first][tmp.second].id].c_str());
flag = true;
}
printf("\n");
}
}
return 0;
}
时间： 2024-12-22 02:09:03