紫書-第五章-例題

衹爲總結，不爲其他，敲過代碼，但也可能會遺忘，權當加深印象了；

5-1 大理石在哪裏（UVa 10474)

很容易的題目，sort+lower_bound，不過自己對lower_bound還是用的很少，欸，所以還是不太敢用的= =；

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 10005;
int num[maxn];
int main() {
    int n, q, kase = 0;
    while (cin >> n >> q&&n) {
        printf("CASE# %d:\n", ++kase);
        for (int i = 0; i < n; i++)
            scanf("%d", &num[i]);
        sort(num, num + n);
        int temp;
        while(q--){
            scanf("%d", &temp);
            int k = lower_bound(num, num + n, temp) - num;
            if (num[k] == temp&&k<n)
                printf("%d found at %d\n", temp, k + 1);
            else printf("%d not found\n", temp);
        }
    }
    return 0;
}

5-2 木塊問題（UVa 101）

直接看劉老師的題意，還是有點懵逼的，不過再看劉老師給的代碼，就理解了題意，同時又驚嘆于劉老師的代碼，寫得如此優美，很多意想不到的東西；

嗯，這題忘記得差不多了，不過題目不難，看了題目，估計可以隨手寫；

#include<cstdio>
#include<iostream>
#include<vector>
#include<string>
using namespace std;

const int maxn = 30;
int n;
vector<int> pile[maxn];

void find(int a, int& p, int& h) {
    for (p = 0; p < n; p++)
        for (h = 0; h < pile[p].size(); h++)
            if (pile[p][h] == a)
                return;
}

void clear(int p, int h) {
    for (size_t i = h + 1; i < pile[p].size(); i++)
        pile[pile[p][i]].push_back(pile[p][i]);
    pile[p].resize(h + 1);
}

void pile_onto(int p, int h, int p2) {
    for (size_t i = h; i < pile[p].size(); i++)
        pile[p2].push_back(pile[p][i]);
    pile[p].resize(h);
}

int main() {
    int a, b;
    cin >> n;
    string s1, s2;
    for (int i = 0; i < n; i++)
        pile[i].push_back(i);
    while (cin >> s1 >> a >> s2 >> b) {
        if (s1 == "quit") break;
        int pa, pb, ha, hb;
        find(a, pa, ha);
        find(b, pb, hb);
        if (pa == pb) continue;
        if (s2 == "onto") clear(pb, hb);
        if (s1 == "move") clear(pa, ha);
        pile_onto(pa, ha, pb);
    }
    for (int i = 0; i < n; i++) {
        printf("%d:", i);
        for (size_t j = 0; j < pile[i].size(); j++)
            printf(" %d", pile[i][j]);
        putchar('\n');
    }
    return 0;
}

5-3 安迪的第一個字典 （UVa 10815）

這題很簡答，就是把輸入轉換一下格式，然後字符串流，直接扔進set就輸出了；

#include<iostream>
#include<string>
#include<cctype>
#include<set>
#include<sstream>
using namespace std;
set<string> dict;
int main() {
    string s, buf;
    while (cin >> s) {
        for (int i = 0; i < s.size(); i++)
            if (isalpha(s[i])) s[i] = tolower(s[i]);
            else s[i] = ' ';
            stringstream ss(s);
            while (ss >> buf) dict.insert(buf);
    }
    for (set<string>::iterator it = dict.begin(); it != dict.end(); it++)
        cout << *it << endl;
    return 0;
}

5-4反片語 （UVa 156）

額，這題轉換格式再sort；

#include<iostream>
#include<algorithm>
#include<map>
#include<cctype>
#include<vector>
#include<string>
using namespace std;
vector<string> word;
map<string, int> temp;
string reper(string &s) {
    string t = s;
    for (int i = 0; i < s.size(); i++)
        t[i] = tolower(s[i]);
    sort(t.begin(), t.end());
    return t;
}
int main() {
    string s;
    while (cin >> s) {
        if (s[0] == '#') break;
        word.push_back(s);
        string t = reper(s);
        if (!temp.count(t))
            temp[t] = 0;
        temp[t]++;
    }
    vector<string> ans;
    for (vector<string>::iterator it = word.begin(); it != word.end(); it++)
        if (temp[reper(*it)] == 1)
            ans.push_back(*it);
    sort(ans.begin(), ans.end());
    for (int i = 0; i < ans.size(); i++)
        cout << ans[i] << endl;
    return 0;
}

5-5集合棧計算機（UVa 12096)

嗯，這題有點坑啊，讀不太懂劉老師描述的題意，是看了代碼和樣例才搞懂的，總的來説也不難= =；

#include<iostream>
#include<stack>
#include<vector>
#include<set>
#include<string>
#include<map>
#include<algorithm>
#include<iterator>
using namespace std;
typedef set<int> Set;
map<Set, int> idcache;
vector<Set> setcache;
int id(Set x) {
    if (idcache.count(x))
        return idcache[x];
    setcache.push_back(x);
    return idcache[x] = setcache.size() - 1;
}
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
int main() {
    int t;
    cin >> t;
    while (t--) {
        stack<int> s;
        int n;
        cin >> n;
        for (int i = 0; i < n; i++) {
            string op;
            cin >> op;
            if (op[0] == 'P')
                s.push(id(Set()));
            else if (op[0] == 'D')
                s.push(s.top());
            else {
                Set x1 = setcache[s.top()];
                s.pop();
                Set x2 = setcache[s.top()];
                s.pop();
                Set x;
                if (op[0] == 'U')
                    set_union(ALL(x1), ALL(x2), INS(x));
                if (op[0] == 'I')
                    set_intersection(ALL(x1), ALL(x2), INS(x));
                if (op[0] == 'A') {
                    x = x2;
                    x.insert(id(x1));
                }
                s.push(id(x));
            }
            cout << setcache[s.top()].size() << endl;
        }
        cout << "***\n";
    }
    return 0;
}

5-6團體隊列（UVa 540）

嗯，用隊列去模擬優先隊列，也不難；

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<map>
#include<queue>
#include<cstdio>
using namespace std;
const int maxn = 1e3 + 10;
int main() {
    int t, kase=0;
    while (cin >> t&&t) {
        printf("Scenario #%d\n", ++kase);
        map<int, int> team;
        int n, x;
        for (int i = 0; i < t; i++) {
            cin >> n;
            while (n--) {
                scanf("%d", &x);
                team[x] = i;
            }
        }
        queue<int> q, q2[maxn];
        for (;;) {
            char cmd[10];
            scanf("%s", cmd);
            if (cmd[0] == 'S') break;
            else if (cmd[0] == 'D') {
                int t = q.front();
                printf("%d\n", q2[t].front());
                q2[t].pop();
                if (q2[t].empty()) q.pop();
            }
            else if (cmd[0] == 'E') {
                scanf("%d", &x);
                if (q2[team[x]].empty())
                    q.push(team[x]);
                q2[team[x]].push(x);
            }
        }
        printf("\n");
    }
    return 0;
}

5-7丑數（UVa 136）

不難，代碼貌似有點玄，不過還好，想一下就清楚了；

#include<iostream>
#include<map>
#include<vector>
#include<queue>
#include<set>
#include <functional>
using namespace std;
typedef long long ll;
const int temp[3] = { 2,3,5 };
int main() {
    priority_queue<ll, vector<ll>, greater<ll> > pq;
    set<ll> s;
    pq.push(1);
    s.insert(1);
    for (int i = 1;; i++) {
        ll x = pq.top();
        pq.pop();
        if (i == 1500) {
            cout << "The 1500'th ugly number is " << x << ".\n";
            break;
        }
        for (int j = 0; j < 3; j++) {
            ll x2 = x*temp[j];
            if (!s.count(x2)) {
                s.insert(x2);
                pq.push(x2);
            }
        }
    }
    return 0;
}

 

5-8 Unix Is命令 (UVa 400)

 

嗯哼，貌似當初寫這題，折騰好久的呢，控制輸出的時候的行列那些= =額，貌似忘了

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
using namespace std;
const int maxcol = 60;
const int maxn = 105;
string failnames[maxn];
void print(const string&s, int len, char extra) {
    cout << s;
    for (int i = 0; i < len - s.length(); i++)
        cout << extra;
}
int main() {
    int n;
    while (cin>>n) {
        int M = 0;
        for (int i = 0; i < n; i++) {
            cin >> failnames[i];
            M = max(M, (int)failnames[i].length());
        }
        int cols = (maxcol - M) / (M + 2) + 1;
        int rows = (n - 1) / cols + 1;
        print("", maxcol, '-');
        cout << endl;
        sort(failnames, failnames + n);
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                int idx = j*rows + i;
                if (idx < n) print(failnames[idx], j == cols - 1 ? M : M + 2, ' ');
            }
            cout << endl;
        }
    }
    return 0;
}

5-9 數據庫（UVa 1592）

額，這題不難的，不過曲解了劉老師的描述，貌似是看了劉老師的代碼，然後才寫出來的？當時的忘了是TLE了還是WA了= =欸，記性好差額

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<vector>
#include<map>
#include<string>
#include<utility>
#include<cstring>
using namespace std;
const int M = 15;
const int N = 10005;
typedef pair<int, int> P;
map<string,int> idcache;
int cnt, n, m, db[N][M];
string getstr(char * s) {
    int c;
    while ((c = getchar()) == '\n' || c == ' ')
        ;
    s[0] = c;
    for (int i = 1; (c = getchar()) != ',' && c != '\n' && c != EOF; i++)
        s[i] = c;
    string str = s;
    return str;
}
int ID(const string &s) {
    if (!idcache.count(s))
        idcache[s]=++cnt;
    return idcache[s];
}
void find() {
    for (int i = 0; i < m; i++)
        for (int j = i + 1; j < m; j++) {
            map<P, int> ans;
            for (int k = 0; k < n; k++) {
                P p = make_pair(db[k][i], db[k][j]);
                if (ans.count(p)) {
                    cout << "NO\n";
                    cout << ans[p] + 1 << " " << k+1 << endl;
                    cout << i+1 << " " << j+1 << endl;
                    return;
                }
                ans[p] = k;
            }
        }
    cout << "YES\n";
}
int main() {
    while (cin >> n >> m) {
        char str[105];
        idcache.clear();
        cnt = 0;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++) {
                memset(str, 0, sizeof(str));
                db[i][j] = ID(getstr(str));
            }
        find();
    }
    return 0;
}

5-10 PGA巡迴賽的獎金（UVa 207)

這題好麻煩啊= =，代碼貼在另一個隨筆了；

5-11 郵件傳輸代理的交互（UVa 814）

嗯，題意里有個東西還不是很清楚，不過貌似不影響AC，而且劉老師代碼裏面也沒體現

#include<iostream>
#include<string>
#include<set>
#include<vector>
#include<map>
using namespace std;
#define FOR(i, n) for (int i = 0; i < (n);i++)
void get_add(const string& temp, string& user, string& mta) {
    int mid = temp.find('@');
    user = temp.substr(0, mid);
    mta = temp.substr(mid + 1);
}
int main() {
    string temp, str, user1, user2, mta1, mta2;
    set<string> sstr;
    int n;
    while (cin >> temp&&temp != "*") {
        cin >> mta1 >> n;
        FOR(i, n) {
            cin >> temp;
            sstr.insert(temp + "@" + mta1);
        }
    }
    while (cin >> str&&str != "*") {
        get_add(str, user1, mta1);
        //cout << user1 << "  " << mta1 << endl;
        vector<string> mta;
        map<string, vector<string> > dest;
        set<string> vis;
        while (cin >> temp&&temp != "*") {
            get_add(temp, user2, mta2);
            if (vis.count(temp)) continue;
            vis.insert(temp);
            if (!dest.count(mta2)) {
                mta.push_back(mta2);
                dest[mta2] = vector<string>();
            }
            dest[mta2].push_back(temp);
        }
        getline(cin, temp);
        string data;
        while (getline(cin, temp) && temp[0] != '*')
            data += "     " + temp + "\n";
        FOR(i, mta.size()) {
            string mta2 = mta[i];
            vector<string> users = dest[mta2];
            cout << "Connection between " << mta1 << " and " << mta2 << endl;
            cout << "     HELO " << mta1 << endl;
            cout << "     250\n";
            cout << "     MAIL FROM:<" << str << ">\n";
            cout << "     250\n";
            bool ok = false;
            FOR(i, users.size()) {
                cout << "     RCPT TO:<" << users[i] << ">\n";
                if (sstr.count(users[i])) {
                    ok = true;
                    cout << "     250\n";
                }
                else cout << "     550\n";
            }
            if (ok) {
                cout << "     DATA\n     354\n";
                cout << data;
                cout << "     .\n";
                cout << "     250\n";
            }
            cout << "     QUIT\n     221\n";
        }
    }
    return 0;
}

5-12城市正視圖（UVa 221）

嗯，第二次看第五章，之前直接把這個跳過的，後來才發現，其實很簡單而已= =就是把各個X坐標都存起來，然後各個X坐標分析一下能否見，然而自己被搞暈過，看到最後面的從前往後遍歷，忘記了sort了一下，位置已經不然讀入的時候的位置了；

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<set>
using namespace std;

const int maxn = 105;
struct building {
    int id;
    double x, y, d, w, h;
    bool operator < (const building& rhs)const {
        return x < rhs.x || (x == rhs.x &&y < rhs.y);
    }
} b[maxn];
int n;
double x[maxn * 2];
bool cover(int i, double mx) {
    return b[i].x <= mx&&b[i].x + b[i].w >= mx;
}
bool visible(int i, double mx) {
    if (!cover(i, mx)) return false;
    for (int k = 0; k < n; k++)
        if (b[k].y<b[i].y&&b[k].h>=b[i].h&&cover(k, mx)) return false;
    return true;
}
int main() {
    int kase = 0;
    while (scanf("%d", &n) == 1 && n) {
        for (int i = 0; i < n; i++) {
            scanf("%lf%lf%lf%lf%lf", &b[i].x, &b[i].y, &b[i].w, &b[i].d, &b[i].h);
            x[i * 2] = b[i].x;
            x[i * 2 + 1] = b[i].x + b[i].w;
            b[i].id = i + 1;
        }
        sort(b, b + n);
        sort(x, x + 2 * n);
        int m = unique(x, x + n * 2) - x;
        if (kase++) printf("\n");
        printf("For map #%d, the visible buildings are numbered as follows:\n%d", kase, b[0].id);
        for (int i = 1; i < n; i++) {
            bool vis = false;
            for (int j = 0; j < m - 1; j++)
                if (visible(i, (x[j] + x[j + 1]) / 2)) {
                    vis = true;
                    break;
                }
            if (vis) printf(" %d", b[i].id);
        }
        printf("\n");
    }
    return 0;
}