[刷题]算法竞赛入门经典(第2版) 5-10/UVa1597 - Searching the Web
题意:不难理解,照搬题意的解法。
代码:(Accepted,0.190s)
//UVa1597 - Searching the Web
//#define _XIENAOBAN_
#include<iostream>
#include<sstream>
#include<cstring>
#include<string>
#include<vector>
#include<map>
#include<set>
using namespace std;
map<string, set<unsigned> > word;
char lines[1510][85];
unsigned N, M, la[105], ln(1);//la:last line of the article; ln:line number
char tmpword[80], tmpline[88], w1[80], w2[80], w3[80];
void ONLY(bool& flag, const string& on) {
    auto &t(word[on]);
    auto p(t.begin());
    for (unsigned i(1);i <= N;++i) {
        while (p != t.end() && *p < la[i - 1]) ++p;
        if (p != t.end() && *p < la[i]) {
            if (flag) puts("----------");
            else flag = true;
            while (p != t.end() && *p < la[i])
                puts(lines[*p++]);
        }
    }
}
void NOT(bool& flag, const string& on) {
    auto &t(word[on]);
    auto p(t.begin());
    for (unsigned i(0);i < N;++i) {
        while (p != t.end() && *p < la[i]) ++p;
        if (p == t.end() || *p >= la[i + 1]) {
            if (flag) puts("----------");
            else flag = true;
            for (auto j(la[i]);j < la[i + 1];++j)
                puts(lines[j]);
        }
    }
}
void AND(bool& flag, const string& le, const string& ri) {
    auto &t1(word[le]), &t2(word[ri]);
    auto p1(t1.begin()),
        p2(t2.begin());
    for (unsigned i(1);i <= N;++i) {
        while (p1 != t1.end() && *p1 < la[i - 1]) ++p1;
        while (p2 != t2.end() && *p2 < la[i - 1]) ++p2;
        if ((p1 != t1.end() && *p1 < la[i]) &&
            (p2 != t2.end() && *p2 < la[i])) {
            if (flag) puts("----------");
            else flag = true;
            unsigned re;
            while ((p1 != t1.end() && *p1 < la[i]) ||
                (p2 != t2.end() && *p2 < la[i])) {
                if (p1 == t1.end() || *p1 >= la[i]) re = *p2++;
                else if (p2 == t2.end() || *p2 >= la[i]) re = *p1++;
                else if (*p2 > *p1) re = *p1++;
                else if (*p2 < *p1) re = *p2++;
                else re = *p1++, *p2++;
                puts(lines[re]);
            }
        }
    }
}
void OR(bool& flag, const string& le, const string& ri) {
    auto p1(word[le].begin()),
        p2(word[ri].begin());
    for (unsigned i(1);i <= N;++i) {
        while (p1 != word[le].end() && *p1 < la[i - 1]) ++p1;
        while (p2 != word[ri].end() && *p2 < la[i - 1]) ++p2;
        if ((p1 != word[le].end() && *p1 < la[i]) ||
            (p2 != word[ri].end() && *p2 < la[i])) {
            if (flag) puts("----------");
            else flag = true;
            unsigned re;
            while ((p1 != word[le].end() && *p1 < la[i]) ||
                (p2 != word[ri].end() && *p2 < la[i])) {
                if (p1 == word[le].end() || *p1 >= la[i]) re = *p2++;
                else if (p2 == word[ri].end() || *p2 >= la[i]) re = *p1++;
                else if (*p2 > *p1) re = *p1++;
                else if (*p2 < *p1) re = *p2++;
                else re = *p1++, *p2++;
                puts(lines[re]);
            }
        }
    }
}
int main()
{
#ifdef _XIENAOBAN_
#define gets(T) gets_s(T, 88)
    freopen("in.txt", "r", stdin);
#endif
    scanf("%d", &N);
    getchar();
    for (unsigned now(0);now < N;++now) {                       //read contexts of articles
        la[now] = ln;
        while (gets(tmpline) && strcmp(tmpline, "**********")) {
            strcpy(lines[ln], tmpline);
            char *pl(tmpline);
            do {
                if ((*pl >= 'a' && *pl <= 'z') || (*pl >= 'A' && *pl <= 'Z')) {
                    char *pw(tmpword);
                    while ((*pl >= 'a' && *pl <= 'z') || (*pl >= 'A' && *pl <= 'Z')) {
                        if (*pl >= 'A' && *pl <= 'Z') *pl += 32;
                        *pw++ = *pl++;
                    }
                    *pw = 0;
                    word[tmpword].insert(ln);
                }
            } while (*pl++ != 0);
            ++ln;
        }
    }
    la[N] = ln;
    scanf("%d", &M);
    getchar();
    int nnnn = 0;
    while (M--) {                                               //read requests
        gets(tmpline);
        char *p = tmpline, *pw1 = w1, *pw2 = w2, *pw3 = w3;
        while (*p && *p != ' ') *pw1++ = *p++;
        *pw1 = '\0';
        if (*p) ++p;
        while (*p && *p != ' ') *pw2++ = *p++;
        *pw2 = '\0';
        if (*p) ++p;
        while (*p && *p != ' ') *pw3++ = *p++;
        *pw3 = '\0';
        bool flag(false);
        if (!*w2)               ONLY(flag, w1);
        else if (*w1 == 'N')    NOT(flag, w2);
        else if (*w2 == 'A')    AND(flag, w1, w3);
        else                    OR(flag, w1, w3);
        if (!flag) puts("Sorry, I found nothing.");
        puts("==========");
        //printf("==========%d\n", ++nnnn);//
    }
    return 0;
}分析:我选择死亡。。。难倒是不难,很快就解出来了。但是一开始用了近1秒的时间,很郁闷。于是又花了几天的时间去研究别人的代码,也包括把之前sstream全改掉用gets和puts,把only not and or全部分开做函数,写的老长。用gets和puts的版本用了0.600s,还是比别人的都慢好多。GitHub上看到大神0.080s的(https://github.com/morris821028/UVa/blob/master/temp/1597%20-%20Searching%20the%20Web.cpp),我天这是差了多少倍!于是仿着大神的代码也写了个,结果1.090s!!??一定是我没抄到其精髓?!花了几天时间了,结果还变慢了,好难过啊。。。然后过了几天我又回来研究了。我发现其实网上看到的算法全都是大同小异的,那我的代码一定是哪里有问题。看到大神函数里面的这一句:set<int> &t = wordInLine[args[1]]; 一开始我以他只是图个书写方便,但想了想,这样就不需要每次寻找wordInLine[args[1]]的值了,于是我就抱着试试的心态,也加了句类似的。结果瞬间从0.960s降到0.190s!新技能get! 
下面是仅仅没有给word[on]、word[le]、word[ri]加引用的版本。 
代码:(Accepted,1.090s)
//UVa1597 - Searching the Web
//#define _XIENAOBAN_
#include<iostream>
#include<sstream>
#include<cstring>
#include<string>
#include<vector>
#include<map>
#include<set>
using namespace std;
map<string, set<unsigned> > word;
char lines[1510][85];
unsigned N, M, la[105], ln(1);//la:last line of the article; ln:line number
char tmpword[80], tmpline[88], w1[80], w2[80], w3[80];
void ONLY(bool& flag, const string& on) {
    auto p(word[on].begin());
    for (unsigned i(1);i <= N;++i) {
        while (p != word[on].end() && *p < la[i - 1]) ++p;
        if (p != word[on].end() && *p < la[i]) {
            if (flag) puts("----------");
            else flag = true;
            while (p != word[on].end() && *p < la[i])
                puts(lines[*p++]);
        }
    }
}
void NOT(bool& flag, const string& on) {
    auto p(word[on].begin());
    for (unsigned i(0);i < N;++i) {
        while (p != word[on].end() && *p < la[i]) ++p;
        if (p == word[on].end() || *p >= la[i + 1]) {
            if (flag) puts("----------");
            else flag = true;
            for (auto j(la[i]);j < la[i + 1];++j)
                puts(lines[j]);
        }
    }
}
void AND(bool& flag, const string& le, const string& ri) {
    auto p1(word[le].begin()),
        p2(word[ri].begin());
    for (unsigned i(1);i <= N;++i) {
        while (p1 != word[le].end() && *p1 < la[i - 1]) ++p1;
        while (p2 != word[ri].end() && *p2 < la[i - 1]) ++p2;
        if ((p1 != word[le].end() && *p1 < la[i]) &&
            (p2 != word[ri].end() && *p2 < la[i])) {
            if (flag) puts("----------");
            else flag = true;
            unsigned re;
            while ((p1 != word[le].end() && *p1 < la[i]) ||
                (p2 != word[ri].end() && *p2 < la[i])) {
                if (p1 == word[le].end() || *p1 >= la[i]) re = *p2++;
                else if (p2 == word[ri].end() || *p2 >= la[i]) re = *p1++;
                else if (*p2 > *p1) re = *p1++;
                else if (*p2 < *p1) re = *p2++;
                else re = *p1++, *p2++;
                puts(lines[re]);
            }
        }
    }
}
void OR(bool& flag, const string& le, const string& ri) {
    auto p1(word[le].begin()),
        p2(word[ri].begin());
    for (unsigned i(1);i <= N;++i) {
        while (p1 != word[le].end() && *p1 < la[i - 1]) ++p1;
        while (p2 != word[ri].end() && *p2 < la[i - 1]) ++p2;
        if ((p1 != word[le].end() && *p1 < la[i]) ||
            (p2 != word[ri].end() && *p2 < la[i])) {
            if (flag) puts("----------");
            else flag = true;
            unsigned re;
            while ((p1 != word[le].end() && *p1 < la[i]) ||
                (p2 != word[ri].end() && *p2 < la[i])) {
                if (p1 == word[le].end() || *p1 >= la[i]) re = *p2++;
                else if (p2 == word[ri].end() || *p2 >= la[i]) re = *p1++;
                else if (*p2 > *p1) re = *p1++;
                else if (*p2 < *p1) re = *p2++;
                else re = *p1++, *p2++;
                puts(lines[re]);
            }
        }
    }
}
int main()
{
#ifdef _XIENAOBAN_
#define gets(T) gets_s(T, 88)
    freopen("in.txt", "r", stdin);
#endif
    scanf("%d", &N);
    getchar();
    for (unsigned now(0);now < N;++now) {                       //read contexts of articles
        la[now] = ln;
        while (gets(tmpline) && strcmp(tmpline, "**********")) {
            strcpy(lines[ln], tmpline);
            char *pl(tmpline);
            do {
                if ((*pl >= 'a' && *pl <= 'z') || (*pl >= 'A' && *pl <= 'Z')) {
                    char *pw(tmpword);
                    while ((*pl >= 'a' && *pl <= 'z') || (*pl >= 'A' && *pl <= 'Z')) {
                        if (*pl >= 'A' && *pl <= 'Z') *pl += 32;
                        *pw++ = *pl++;
                    }
                    *pw = 0;
                    word[tmpword].insert(ln);
                }
            } while (*pl++ != 0);
            ++ln;
        }
    }
    la[N] = ln;
    scanf("%d", &M);
    getchar();
    int nnnn = 0;
    while (M--) {                                               //read requests
        gets(tmpline);
        char *p = tmpline, *pw1 = w1, *pw2 = w2, *pw3 = w3;
        while (*p && *p != ' ') *pw1++ = *p++;
        *pw1 = '\0';
        if (*p) ++p;
        while (*p && *p != ' ') *pw2++ = *p++;
        *pw2 = '\0';
        if (*p) ++p;
        while (*p && *p != ' ') *pw3++ = *p++;
        *pw3 = '\0';
        bool flag(false);
        if (!*w2)               ONLY(flag, w1);
        else if (*w1 == 'N')    NOT(flag, w2);
        else if (*w2 == 'A')    AND(flag, w1, w3);
        else                    OR(flag, w1, w3);
        if (!flag) puts("Sorry, I found nothing.");
        puts("==========");
        //printf("==========%d\n", ++nnnn);//
    }
    return 0;
}哎呀呀,就少那么几行,天壤之别。顺便把一开始做的版本也贴了:
代码:(Accepted,0.960s)
//UVa1597 - Searching the Web
#include<iostream>
#include<sstream>
#include<string>
#include<vector>
#include<map>
#include<set>
using namespace std;
struct article {
    map<string, vector<unsigned> > word;
    vector<string> line;
};
const article tmpart;//an empty article,just for push_back
unsigned N, M;
string tmpstr;
vector<article> dat;
inline bool ONLY(const article& ar, const string& l, const string& r) {
    return ar.word.count(l);
}
inline bool AND(const article& ar, const string& l, const string& r) {
    return ar.word.count(l) && ar.word.count(r);
}
inline bool OR(const article& ar, const string& l, const string& r) {
    return ar.word.count(l) || ar.word.count(r);
}
void NOT(const string& on) {
    bool flag(false);
    for (const auto& r : dat) {
        if (!r.word.count(on)) {
            if (flag) cout << "----------\n";
            else flag = true;
            for (const auto& rr : r.line)
                cout << rr << '\n';
        }
    }
    if (!flag) cout << "Sorry, I found nothing.\n";
}
void solve(bool(*f)(const article&, const string&, const string&), const string& le, const string& ri) {
    bool flag(false);
    for (auto& r : dat) {//why it can't be "const auto& r"? because if a map is const,r.word[i] would be invalid
        if (f(r, le, ri)) {
            set<unsigned> re;
            if (r.word.count(le)) for (const auto& rr : r.word[le]) re.insert(rr);
            if (r.word.count(ri)) for (const auto& rr : r.word[ri]) re.insert(rr);
            if (re.empty()) continue;
            if (flag) cout << "----------\n";
            else flag = true;
            for (const auto& rr : re)
                cout << r.line[rr] << '\n';
        }
    }
    if (!flag) cout << "Sorry, I found nothing.\n";
}
int main()
{
    //freopen("in.txt", "r", stdin);//
    std::ios::sync_with_stdio(false);
    cin >> N;
    cin.ignore();//absorb '\n'
    for (unsigned now(0);now < N;++now) {                       //read contexts of articles
        int ln(0);//line number of current article
        dat.push_back(tmpart);//create a empty article
        while (getline(cin, tmpstr) && tmpstr != "**********") {
            dat[now].line.push_back(tmpstr);
            for (auto& r : tmpstr) {
                if (r >= 'a'&&r <= 'z') continue;
                if (r >= 'A'&&r <= 'Z') r += 32;
                else r = ' ';
            }
            istringstream in(tmpstr);
            while (in >> tmpstr)
                dat[now].word[tmpstr].push_back(ln);
            ++ln;
        }
    }
    cin >> M;
    cin.ignore();
    while (M--) {                                               //read requests
        string sl, sm, sr;//left middle right
        getline(cin, tmpstr);
        istringstream in(tmpstr);
        in >> sl;
        if (!(in >> sm))            solve(ONLY, sl, sl);
        else if (!(in >> sr))       NOT(sm);//why are you so special
        else if (sm[0] == 'A')      solve(AND, sl, sr);
        else                        solve(OR, sl, sr);
        cout << "==========\n";
    }
    return 0;
}[刷题]算法竞赛入门经典(第2版) 5-10/UVa1597 - Searching the Web的更多相关文章
- [刷题]算法竞赛入门经典(第2版) 4-6/UVa508 - Morse Mismatches
		书上具体所有题目:http://pan.baidu.com/s/1hssH0KO 代码:(Accepted,10 ms) //UVa508 - Morse Mismatches #include< ... 
- [刷题]算法竞赛入门经典(第2版) 5-15/UVa12333 - Revenge of Fibonacci
		题意:在前100000个Fibonacci(以下简称F)数字里,能否在这100000个F里找出以某些数字作为开头的F.要求找出下标最小的.没找到输出-1. 代码:(Accepted,0.250s) / ... 
- [刷题]算法竞赛入门经典(第2版) 5-13/UVa822 - Queue and A
		题意:模拟客服MM,一共有N种话题,每个客服MM支持处理其中的i个(i < N),处理的话题还有优先级.为了简化流程方便出题,设每个话题都是每隔m分钟来咨询一次.现知道每个话题前来咨询的时间.间 ... 
- [刷题]算法竞赛入门经典(第2版) 4-5/UVa1590 - IP Networks
		书上具体所有题目:http://pan.baidu.com/s/1hssH0KO 代码:(Accepted,0 ms) //UVa1590 - IP Networks #include<iost ... 
- [刷题]算法竞赛入门经典(第2版) 6-7/UVa804 - Petri Net Simulation
		题意:模拟Petri网的执行.虽然没听说过Petri网,但是题目描述的很清晰. 代码:(Accepted,0.210s) //UVa804 - Petri Net Simulation //Accep ... 
- [刷题]算法竞赛入门经典(第2版) 6-6/UVa12166 - Equilibrium Mobile
		题意:二叉树代表使得平衡天平,修改最少值使之平衡. 代码:(Accepted,0.030s) //UVa12166 - Equilibrium Mobile //Accepted 0.030s //# ... 
- [刷题]算法竞赛入门经典(第2版) 6-1/UVa673 6-2/UVa712 6-3/UVa536
		这三题比较简单,只放代码了. 题目:6-1 UVa673 - Parentheses Balance //UVa673 - Parentheses Balance //Accepted 0.000s ... 
- [刷题]算法竞赛入门经典(第2版) 5-16/UVa212 - Use of Hospital Facilities
		题意:模拟患者做手术. 其条件为:医院有Nop个手术室.准备手术室要Mop分钟,另有Nre个恢复用的床.准备每张床要Mre分钟,早上Ts点整医院开张,从手术室手术完毕转移到回复床要Mtr分钟.现在医院 ... 
- [刷题]算法竞赛入门经典(第2版) 5-11/UVa12504 - Updating a Dictionary
		题意:对比新老字典的区别:内容多了.少了还是修改了. 代码:(Accepted,0.000s) //UVa12504 - Updating a Dictionary //#define _XieNao ... 
随机推荐
- 利用shell实现批量添加用户
			批量添加用户并设置随机密码,把添加的用户的名字和密码保存到文件中. [root@lamp scripts]# cat user.sh #!/bin/sh ` do pass=$(-) //取随机数的方 ... 
- OC--Runtime知识点整理
			1.Runtime简介 因为Objc是一门动态语言,所以它总是想办法把一些决定工作从编译连接推迟到运行时.也就是说只有编译器是不够的,还需要一个运行时系统 (runtime system) 来执行编译 ... 
- 【Egret】Lakeshore 使用中的一些疑难解决技巧!
			用Lakeshore 1.2.1版本发布的html,会出现一些用户不想要的东西,下面讲讲如何去掉: 一.问题:游戏或者动画在PC端也能跟随游览器自适应. 解决方法:①找到发布文件下的 egret_l ... 
- Html5-测试Canvas
			// 浏览器不支持Html5 Canvas"; var theCanvas=document.getElementById("canvas_one"); if(!th ... 
- 什么是测试开发工程师-google的解释
			什么是测试开发工程师-google的解释 “ 软件测试开发工程师[SET or Software Engineer in Test],和软件开发工程师一样是开发工程师,主要负责软件的可测试性.他们参与 ... 
- android 开机自启动实现
			App的开机自启动可以通过注册广播接收器接收开机广播来实现,具体步骤如下: 1.创建 BroadcastReceiver 的派生类,并重写 onReceive() 函数: /** * Created ... 
- Android IPC机制全解析<一>
			概要 多进程概念及多进程常见注意事项 IPC基础:Android序列化和Binder 跨进程常见的几种通信方式:Bundle通过Intent传递数据,文件共享,ContentProvider,基于Bi ... 
- Android -- 贝塞尔二阶实现饿了么加入购物车效果
			1,上周我们实现了简单的三阶贝塞尔曲线效果实例,今天是使用二阶贝塞尔曲线加动画实现的加入购物车效果,在码代码过程中出现了些问题,过一下和大家来探讨探讨,先看一下效果图 2,从上面的效果来看我们基本上可 ... 
- C#基础语法(一)
			一.基础语法 1.C#区分大小写,所以myVar和MyVar是两个不同的变量. 2.每个C#可执行文件(如控制台应用程序,Windows应用程序和Windows服务)都必须有一个入口点----Main ... 
- es6基础系列二:Number
			es6中关于Number类型的变化,主要有以下几点 Number.isInteger(新增,判断是否为整数) Number.EPSILON(新增,判断是否可忽略的误差) Number.MAX_SAFE ... 
