uvalive 3942(dp + trie)
2015-04-09 00:07
267 查看
题意:有一个长字符串,然后给出了n个不同的单词,问长字符串是由这些单词构成的不同方案有几种。
题解:明显是dp,f[i]表示从字符i开始有多少种方案,f[i] = sum{f[i + len(x)},x是后半段字符串的前缀,如果枚举前缀肯定会超时,那么用前缀树trie可以直接找到所有的前缀x,用在状态转移方程中。
#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;
const int MOD = 20071027;
const int N = 4000 * 102;
int val
, Next
[30], sz, n;
ll f
;
char str[300005], str2[104];
void init() {
memset(Next[0], 0, sizeof(Next[0]));
val[0] = 0;
sz = 1;
}
void insert(char *s) {
int u = 0, len = strlen(s);
for (int i = 0; i < len; i++) {
int k = s[i] - 'a';
if (!Next[u][k]) {
memset(Next[sz], 0, sizeof(Next[sz]));
val[sz] = 0;
Next[u][k] = sz++;
}
u = Next[u][k];
}
val[u] = 1;
}
int main() {
int cas = 1;
while (scanf("%s", str) == 1) {
init();
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%s", str2);
insert(str2);
}
int len = strlen(str);
memset(f, 0, sizeof(f));
f[len] = 1;
for (int i = len - 1; i >= 0; i--) {
int u = 0;
for (int j = i, depth = 1; j < len; j++, depth++) {
int k = str[j] - 'a';
if (Next[u][k]) {
u = Next[u][k];
if (val[u])
f[i] = (f[i] + f[i + depth]) % MOD;
}
else break;
}
}
printf("Case %d: %lld\n", cas++, f[0]);
}
return 0;
}
题解:明显是dp,f[i]表示从字符i开始有多少种方案,f[i] = sum{f[i + len(x)},x是后半段字符串的前缀,如果枚举前缀肯定会超时,那么用前缀树trie可以直接找到所有的前缀x,用在状态转移方程中。
#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;
const int MOD = 20071027;
const int N = 4000 * 102;
int val
, Next
[30], sz, n;
ll f
;
char str[300005], str2[104];
void init() {
memset(Next[0], 0, sizeof(Next[0]));
val[0] = 0;
sz = 1;
}
void insert(char *s) {
int u = 0, len = strlen(s);
for (int i = 0; i < len; i++) {
int k = s[i] - 'a';
if (!Next[u][k]) {
memset(Next[sz], 0, sizeof(Next[sz]));
val[sz] = 0;
Next[u][k] = sz++;
}
u = Next[u][k];
}
val[u] = 1;
}
int main() {
int cas = 1;
while (scanf("%s", str) == 1) {
init();
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%s", str2);
insert(str2);
}
int len = strlen(str);
memset(f, 0, sizeof(f));
f[len] = 1;
for (int i = len - 1; i >= 0; i--) {
int u = 0;
for (int j = i, depth = 1; j < len; j++, depth++) {
int k = str[j] - 'a';
if (Next[u][k]) {
u = Next[u][k];
if (val[u])
f[i] = (f[i] + f[i + depth]) % MOD;
}
else break;
}
}
printf("Case %d: %lld\n", cas++, f[0]);
}
return 0;
}
相关文章推荐
- UVALive 3942 Trie+dp,白书P209
- UVALive 3942 Remember the Word(trie + dp)
- UVALive - 3942 Remember the Word(trie + dp)
- UVALive - 3942 Remember the Word[Trie DP]
- UVALive 3942 Remember the Word(字典树+dp)
- UVALive - 3942(前缀树DP)
- uvalive 3942 Remember the Word (字典树+DP)
- uvalive 3942 - Remember the Word(Trie)
- UVALive 3942 Remember the Word(字典树 + 简单dp)
- UVALive 3942 Remember the Word 前缀树Trie
- LA 3942 && UVa 1401 Remember the Word (Trie + DP)
- (trie)UVALive - 3942 Remember the Word
- UVALive 3942 - Remember the Word(DP,数组Trie+指针Trie)
- UVALive - 3942 Remember the Word Trie
- UVALive 3942 - Remember the Word(DP,字典树)
- UVALive 5790 Ball Stacking dp
- UVALive 7392 Bundles of Joy【bitset】【类树形DP】【杂题】
- Wooden Signs uvalive gym 区间dp
- UVALive 4015 Caves--树形dp
- uvalive7271(A Math Problem) 数位dp