【BZOJ】1622: [Usaco2008 Open]Word Power 名字的能量(dp/-模拟)
http://www.lydsy.com/JudgeOnline/problem.php?id=1622
这题我搜的题解是dp,我也觉得是dp,但是好像比模拟慢啊!!!!
1400ms不科学!
设f[i][j]为名字i位置的j字母最早出现的位置(向后)
则
f[i][j]=f[i+1][j]
f[i][a[i+1]]=i+1
那么就可以递推出,然后查找即可。。
但是大量的memcpy导致很慢啊。。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }
const int N=1005;
int f[N][265], n, m, ans;
char a[N][N], md[105][40];
int main() {
read(n); read(m);
for1(i, 1, n) { scanf("%s", a[i]+1); for1(j, 1, strlen(a[i]+1)-1) if(a[i][j]>='A'&&a[i][j]<='Z') a[i][j]=a[i][j]-'A'+'a'; }
for1(i, 1, m) { scanf("%s", md[i]); for1(j, 0, strlen(md[i])-1) if(md[i][j]>='A'&&md[i][j]<='Z') md[i][j]=md[i][j]-'A'+'a'; }
for1(i, 1, n) {
CC(f, 0);
for3(j, strlen(a[i]+1)-1, 0) {
memcpy(f[j], f[j+1], sizeof(f[j]));
f[j][(int)a[i][j+1]]=j+1;
}
ans=m;
for1(j, 1, m) {
int t=0;
for2(k, 0, strlen(md[j])) {
t=f[t][(int)md[j][k]];
if(!t) { --ans; break; }
}
}
printf("%d\n", ans);
}
return 0;
}
Description
约 翰想要计算他那N(1≤N≤1000)只奶牛的名字的能量.每只奶牛的名字由不超过1000个字待构成,没有一个名字是空字体串, 约翰有一张“能量字 符串表”,上面有M(1≤M≤100)个代表能量的字符串.每个字符串由不超过30个字体构成,同样不存在空字符串.一个奶牛的名字蕴含多少个能量字符 串,这个名字就有多少能量.所谓“蕴含”,是指某个能量字符串的所有字符都在名字串中按顺序出现(不一定一个紧接着一个).
所有的大写字母和小写字母都是等价的.比如,在贝茜的名字“Bessie”里,蕴含有“Be”
“sI”“EE”以及“Es”等等字符串,但不蕴含“lS”或“eB”.请帮约翰计算他的奶牛的名字的能量.
Input
第1行输入两个整数N和M,之后N行每行输入一个奶牛的名字,之后M行每行输入一个能量字符串.
Output
一共N行,每行一个整数,依次表示一个名字的能量.
Sample Input
5 3
Bessie
Jonathan
Montgomery
Alicia
Angola
se
nGo
Ont
INPUT DETAILS:
There are 5 cows, and their names are "Bessie", "Jonathan",
"Montgomery", "Alicia", and "Angola". The 3 good strings are "se",
"nGo", and "Ont".
Bessie
Jonathan
Montgomery
Alicia
Angola
se
nGo
Ont
INPUT DETAILS:
There are 5 cows, and their names are "Bessie", "Jonathan",
"Montgomery", "Alicia", and "Angola". The 3 good strings are "se",
"nGo", and "Ont".
Sample Output
1
1
2
0
1
OUTPUT DETAILS:
"Bessie" contains "se", "Jonathan" contains "Ont", "Montgomery" contains
both "nGo" and "Ont", Alicia contains none of the good strings, and
"Angola" contains "nGo".
1
2
0
1
OUTPUT DETAILS:
"Bessie" contains "se", "Jonathan" contains "Ont", "Montgomery" contains
both "nGo" and "Ont", Alicia contains none of the good strings, and
"Angola" contains "nGo".
HINT
Source
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