[USACO 2011 Dec Gold] Threatening Letter【后缀】
Problem 3: Threatening Letter [J. Kuipers, 2002] FJ has had a terrible fight with his neighbor and wants to send him
a nasty letter, but wants to remain anonymous. As so many before
him have done, he plans to cut out printed letters and paste them
onto a sheet of paper. He has an infinite number of the most recent
issue of the Moo York Times that has N (1 <= N <= 50,000) uppercase
letters laid out in a long string (though read in as a series of
shorter strings). Likewise, he has a message he'd like to compose
that is a single long string of letters but that is read in as a
set of shorter strings. Being lazy, he wants to make the smallest possible number of cuts.
FJ has a really great set of scissors that enables him to remove
any single-line snippet from the Moo York Times with one cut. He
notices that he can cut entire words or phrases with a single cut,
thus reducing his total number of cuts. What is the minimum amount of cuts he has to make to construct his
letter of M (1 <= M <= 50,000) letters? It is guaranteed that it is possible for FJ to complete his task. Consider a 38 letter Moo York Times: THEQUICKBROWNFOXDO
GJUMPSOVERTHELAZYDOG from which FJ wants to construct a 9 letter message: FOXDOG
DOG These input lines represent a pair of strings: THEQUICKBROWNFOXDOGJUMPSOVERTHELAZYDOG
FOXDOGDOG Since "FOXDOG" exists in the newspaper, FJ can cut this piece out
and then get the last "DOG" by cutting out either instance of the
word "DOG". Thus, he requires but two cuts. PROBLEM NAME: letter INPUT FORMAT: * Line 1: Two space-separated integers: N and M * Lines 2..?: N letters laid out on several input lines; this is the
text of the one copy of the Moo York Times. Each line will
have no more than 80 characters. * Lines ?..?: M letters that are the text of FJ's letter. Each line
will have no more than 80 characters. SAMPLE INPUT (file letter.in): 38 9
THEQUICKBROWNFOXDO
GJUMPSOVERTHELAZYDOG
FOXDOG
DOG OUTPUT FORMAT: * Line 1: The minimum number of cuts FJ has to make to create his
message SAMPLE OUTPUT (file letter.out): 2
一看跟子串相关,就是后缀那一套了。想法是这样的,尽量在文本串中找到更长的子串与当前串匹配。若无法继续匹配了,则重新匹配,答案+1。这里我选择了后缀自动机,实现起来好写。
#include <cstdio>
#include <cstring> const int maxn = 50005; int n, m, ans, now;
int sam[maxn << 1][26], len[maxn << 1], link[maxn << 1], sz, last, p, q, cur, clone;
char ch; int main(void) {
freopen("letter.in", "r", stdin);
freopen("letter.out", "w", stdout);
scanf("%d%d", &n, &m);
link[sz++] = -1;
for (int i = 1; i <= n; ++i) {
while ((ch = getchar()) < 'A');
cur = sz++;
len[cur] = len[last] + 1;
for (p = last; p != -1 && !sam[p][ch - 'A']; p = link[p]) {
sam[p][ch - 'A'] = cur;
}
if (p != -1) {
q = sam[p][ch - 'A'];
if (len[p] + 1 == len[q]) {
link[cur] = q;
}
else {
clone = sz++;
memcpy(sam[clone], sam[q], sizeof sam[0]);
link[clone] = link[q];
len[clone] = len[p] + 1;
for (; p != -1 && sam[p][ch - 'A'] == q; p = link[p]) {
sam[p][ch - 'A'] = clone;
}
link[q] = link[cur] = clone;
}
}
last = cur;
} for (int i = 1; i <= m; ++i) {
while ((ch = getchar()) < 'A');
now = sam[now][ch - 'A'];
if (!now) {
++ans;
now = sam[0][ch - 'A'];
}
}
if (now) {
++ans;
}
printf("%d\n", ans);
return 0;
}
这也算是SAM的模版了叭。
[USACO 2011 Dec Gold] Threatening Letter【后缀】的更多相关文章
- [USACO 2011 Dec Gold] Cow Calisthenics【二分】
Problem 1: Cow Calisthenics [Michael Cohen, 2010] Farmer John continues his never-ending quest to ke ...
- [USACO 2017 Dec Gold] Tutorial
Link: USACO 2017 Dec Gold 传送门 A: 为了保证复杂度明显是从终结点往回退 结果一开始全在想优化建边$dfs$……其实可以不用建边直接$multiset$找可行边跑$bfs$ ...
- [USACO 2011 Nov Gold] Cow Steeplechase【二分图】
传送门:http://www.usaco.org/index.php?page=viewproblem2&cpid=93 很容易发现,这是一个二分图的模型.竖直线是X集,水平线是Y集,若某条竖 ...
- [USACO 2011 Nov Gold] Above the Median【逆序对】
传送门:http://www.usaco.org/index.php?page=viewproblem2&cpid=91 这一题我很快的想出了,把>= x的值改为1,< x的改为- ...
- [Poj3261] [Bzoj1717] [后缀数组论文例题,USACO 2006 December Gold] Milk Patterns [后缀数组可重叠的k次最长重复子串]
和上一题(POJ1743,上一篇博客)相似,只是二分的判断条件是:是否存在一段后缀的个数不小于k #include <iostream> #include <algorithm> ...
- Usaco 2010 Dec Gold Exercise(奶牛健美操)
/*codevs 3279 二分+dfs贪心检验 堆版本 re一个 爆栈了*/ #include<cstdio> #include<queue> #include<cst ...
- BZOJ1774[USACO 2009 Dec Gold 2.Cow Toll Paths]——floyd
题目描述 跟所有人一样,农夫约翰以着宁教我负天下牛,休叫天下牛负我的伟大精神,日日夜夜苦思生 财之道.为了发财,他设置了一系列的规章制度,使得任何一只奶牛在农场中的道路行走,都 要向农夫约翰上交过路费 ...
- BZOJ1775[USACO 2009 Dec Gold 3.Video Game Troubles]——DP
题目描述 输入 * 第1行: 两个由空格隔开的整数: N和V * 第2到第N+1行: 第i+1行表示第i种游戏平台的价格和可以在这种游戏平台上面运行的游 戏.包含: P_i, G_i还有G_i对由空格 ...
- [USACO 2016 Dec Gold] Tutorial
Link: 传送门 A: 贪心从小到大插入,用并查集维护连通性 #include <bits/stdc++.h> using namespace std; #define X first ...
随机推荐
- django 简易博客开发 5 markdown支持、代码高亮、gravatar头像服务
上一篇博客介绍了comments库使用及ajax支持,现在blog已经具备了基本的功能,但是只能发表文字,不支持富文本编辑.今天我们利用markdown添加富文本支持. markdown语法说明: h ...
- Codeforces 344B Simple Molecules
#include<bits/stdc++.h> using namespace std; int main() { int a,b,c; scanf("%d%d%d", ...
- Project Euler:Problem 61 Cyclical figurate numbers
Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are all figurate (polygon ...
- android 自己主动拒接后再取消自己主动拒接,该联系人来电界面无图标显示,且点击挂断无反应
1. 设置一个联系人为自己主动拒接 2. 该联系人来电 3. 取消该联系人的自己主动拒接 4. 该联系人来电 Error: 来电界面无头像显示,直接显示黑屏,且点击拒接butt ...
- 李洪强iOS开发之函数式 编程初窥
函数式 编程初窥 最近在学习Erlang和Python.Erlang是完全的函数式编程语言,Python语言是面向对象的语言,但是它的语法引入了大量的函数式编程思想.越研究越觉得函数式的编程思路可 ...
- 一个很小的C++写的MVC的例子
#include<iostream> #include<vector> //get namespace related stuff using std::cin; using ...
- URL传参中文乱码的一种解决方法
中文乱码是由于,发送和接收方使用的编码解码格式不一致导致,以下是关于url传参解决中文乱码的一种方法,最后根据各种编码格式尝试解码,发现正确的解码格式 string strQueryString = ...
- python itertools
1 product 1.1 一个generator函数 因此它的返回值是一个iterator,可以用for遍历. 1.2 计算product的参数分类 1.2.1 dict和list 只用了dict的 ...
- [BZOJ2144]国家集训队 跳跳棋
题目描述 跳跳棋是在一条数轴上进行的.棋子只能摆在整点上.每个点不能摆超过一个棋子. 我们用跳跳棋来做一个简单的游戏:棋盘上有3颗棋子,分别在a,b,c这三个位置.我们要通过最少的跳动把他们的位置移动 ...
- vue学习1
1.<div id="app">{{message}}<input v-model="message"></div>new ...