Problem A. Pattern Matching

把每个字符串分成第一个之前，最后一个之后，中间的部分 三个部分

每个字符串的中间的部分可以直接拼接

前后两个部分需要判断下是否合法

#include <algorithm>

#include <bitset>

#include <cassert>

#include <cmath>

#include <complex>

#include <cstring>

#include <ctime>

#include <deque>

#include <fstream>

#include <functional>

#include <iomanip>

#include <iostream>

#include <map>

#include <numeric>

#include <queue>

#include <random>

#include <set>

#include <stack>

#include <unordered_map>

#include <unordered_set>

#include <vector>

#define MP make_pair

#define ll long long

#define ld long double

#define null NULL

#define all(a) a.begin(), a.last()

#define forn(i, n) for (int i = 0; i < n; ++i)

#define sz(a) (int)a.size()

#define lson l , m , rt << 1

#define rson m + 1 , r , rt << 1 | 1

#define bitCount(a)  __builtin_popcount(a)

template<class T> int gmax(T &a, T b) { if (b > a) { a = b; return 1; } return 0; }

template<class T> int gmin(T &a, T b) { if (b < a) { a = b; return 1; } return 0; }

using namespace std;

const int INF = 0x3f3f3f3f;

string to_string(string s) { return '"' + s + '"'; }

string to_string(const char* s) { return to_string((string) s); }

string to_string(bool b) { return (b ? "true" : "false"); }

template <typename A, typename B>

string to_string(pair<A, B> p) { return "(" + to_string(p.first) + ", " + to_string(p.second) + ")"; }

template <typename A>

string to_string(A v) { bool first = true; string res = "{"; for (const auto &x : v) { if (!first) { res += ", "; } first = false; res += to_string(x); } res += "}"; return res; }

void debug_out() { cerr << endl; }

template <typename Head, typename... Tail>

void debug_out(Head H, Tail... T) { cerr << " " << to_string(H); debug_out(T...); }

#ifdef LOCAL

#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)

#else

#define debug(...) 42

#endif

char seq[55][105];

int main() {

    int T;

    scanf("%d", &T);

    for(int cas = 1; cas <= T; ++cas) {

        int n;

        scanf("%d", &n);

        for(int i = 0; i < n; ++i) {

            scanf("%s", seq[i]);

        }

        vector<string> front, last;

        string mid_result;

        for(int i = 0; i < n; ++i) {

            int len = strlen(seq[i]);

            vector<string> split;

            string tmp;

            if(seq[i][0] == '*') split.push_back("");

            for(int j = 0; j < len; ++j) {

                // cout << seq[i][j] << endl;

                if(seq[i][j] == '*') {

                    if((int)tmp.size() != 0) split.push_back(tmp);

                    tmp.clear();

                } else tmp += seq[i][j];

            }

            if((int)tmp.size() != 0) split.push_back(tmp);

            // debug(split);

            if(seq[i][len - 1] == '*') split.push_back("");

            front.push_back(split[0]); last.push_back(split.back());

            if(split.size() > 2) {

                for(int j = 1, split_len = split.size(); j < split_len - 1; ++j) {

                    mid_result += split[j];

                }

            }

        }

        auto cmp = [](string &A, string &B) { return A.size() < B.size(); };

        sort(front.begin(), front.end(), cmp);

        sort(last.begin(), last.end(), cmp);

        // debug(front, last, mid_result);

        bool suc = true;

        for(int i = 0, len = front.size(); i < len - 1 && suc; ++i) {

            string &now = front[i]; string &tem = front.back();

            for(int j = 0, len_now = now.size(); j < len_now && suc; ++j) {

                if(now[j] != tem[j]) { suc = false; }

            }

        }

        // debug(suc);

        for(int i = 0, len = last.size(); i < len - 1 && suc; ++i) {

            string &now = last[i]; string &tem = last.back(); int len_tem = tem.size();

            for(int j = 0, len_now = now.size(); j < len_now && suc; ++j) {

                if(now[j] != tem[len_tem + j - len_now]) {

                    // debug(now, tem, j);

                    suc = false;

                }

            }

        }

        string result = front.back() + mid_result + last.back();

        printf("Case #%d: ", cas);

        if(suc == false) printf("*\n");

        else printf("%s\n", result.c_str());

    }

    return 0;

}

/*

2

5

*CONUTS

*COCONUTS

*OCONUTS

*CONUTS

*S

2

*XZ

*XYZ

*/

problem B. Pascal Walk

这是一个 非常巧妙的构造题

首先想到一个杨辉三角的每一个层的和是 2^i ，如果我们可以跳跃就好了，直接按照N的二进制表示，使用对应层的和

问题在于我们无法跳跃，怎么办

通过构造可以发现我们可以将两个很远的两层连在一起，下面举例5，19的情况，具体的逻辑可以去代码里面体会



#include <algorithm>

#include <bitset>

#include <cassert>

#include <cmath>

#include <complex>

#include <cstring>

#include <ctime>

#include <deque>

#include <fstream>

#include <functional>

#include <iomanip>

#include <iostream>

#include <map>

#include <numeric>

#include <queue>

#include <random>

#include <set>

#include <stack>

#include <unordered_map>

#include <unordered_set>

#include <vector>

#define MP make_pair

#define ll long long

#define ld long double

#define null NULL

#define all(a) a.begin(), a.last()

#define forn(i, n) for (int i = 0; i < n; ++i)

#define sz(a) (int)a.size()

#define lson l , m , rt << 1

#define rson m + 1 , r , rt << 1 | 1

#define bitCount(a)  __builtin_popcount(a)

template<class T> int gmax(T &a, T b) { if (b > a) { a = b; return 1; } return 0; }

template<class T> int gmin(T &a, T b) { if (b < a) { a = b; return 1; } return 0; }

using namespace std;

const int INF = 0x3f3f3f3f;

string to_string(string s) { return '"' + s + '"'; }

string to_string(const char* s) { return to_string((string) s); }

string to_string(bool b) { return (b ? "true" : "false"); }

template <typename A, typename B>

string to_string(pair<A, B> p) { return "(" + to_string(p.first) + ", " + to_string(p.second) + ")"; }

template <typename A>

string to_string(A v) { bool first = true; string res = "{"; for (const auto &x : v) { if (!first) { res += ", "; } first = false; res += to_string(x); } res += "}"; return res; }

void debug_out() { cerr << endl; }

template <typename Head, typename... Tail>

void debug_out(Head H, Tail... T) { cerr << " " << to_string(H); debug_out(T...); }

#ifdef LOCAL

#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)

#else

#define debug(...) 42

#endif

void add(vector<pair<int, int> >& vc, int pre, int target, int times) {

    // debug(pre, target, vc);

    int dir = 0;

    if( times % 2 == 0) dir = 0; else dir = 1;

    int preX = -1, preY = 0;

    if(vc.size() != 0) {

        preX = vc.back().first; preY = vc.back().second;

    }

    int needFloor = target - pre - 1;

    if(needFloor == 0) {

        vc.push_back(dir ? MP(preX + 1, preY + 1) : MP(preX + 1, preY));

        for(int i = 0; i < target; ++i) {

            vc.push_back(dir ? MP(vc.back().first, vc.back().second - 1) : MP(vc.back().first, vc.back().second + 1));

        }

    } else {

        vc.push_back(dir ? MP(preX + 1, preY + 1) : MP(preX + 1, preY));

        int count = 2;

        for(int i = 0; i < needFloor - 1; ++i) {

            int tmp_count = count - 1; int now_dir = (i & 1) ^ (needFloor & 1) ^ dir;

            vc.push_back(now_dir ? MP(vc.back().first + 1, vc.back().second + 1) : MP(vc.back().first + 1, vc.back().second));

            while(tmp_count --) {

                vc.push_back(now_dir ? MP(vc.back().first, vc.back().second - 1) : MP(vc.back().first, vc.back().second + 1));

            }

            count ++;

        }

        vc.push_back(dir ? MP(vc.back().first + 1, vc.back().second) : MP(vc.back().first + 1, vc.back().second + 1));

        for(int i = 0; i < target - count + 1; ++i) {

            vc.push_back(dir ? MP(vc.back().first, vc.back().second - 1) : MP(vc.back().first, vc.back().second + 1));

        }

    }

}

int main() {

    int T;

    scanf("%d", &T);

    for(int cas = 1; cas <= T; ++cas) {

        int n;

        scanf("%d", &n);

        vector<pair<int, int> > vc;

        int floor = -1; int cnt = 0; int times = 0;

        while(n) {

            if(n & 1) {

                add(vc, floor, cnt, times);

                times ++;

                floor = cnt;

            }

            cnt ++;

            n /= 2;

        }

        assert((int)vc.size() < 500);

        printf("Case #%d: \n", cas);

        for(int i = 0, len = vc.size(); i < len; ++i) {

            printf("%d %d\n", vc[i].first + 1, vc[i].second + 1);

        }

    }

    return 0;

}

Problem C: Square Dance

这题看起来就是暴力，能过小数据

有个显而易见的优化，就是每次删除一个点之后，下一轮潜在的可能删除点一定是上轮被删点的邻居

复杂度不太会算，题解说这样优化后能到O(R * C)

找邻居这种数据结构 我用十字链表维护的

#include <algorithm>

#include <bitset>

#include <cassert>

#include <cmath>

#include <complex>

#include <cstring>

#include <ctime>

#include <deque>

#include <fstream>

#include <functional>

#include <iomanip>

#include <iostream>

#include <map>

#include <numeric>

#include <queue>

#include <random>

#include <set>

#include <stack>

#include <unordered_map>

#include <unordered_set>

#include <vector>

#define MP make_pair

#define ll long long

#define ld long double

#define null NULL

#define all(a) a.begin(), a.last()

#define forn(i, n) for (int i = 0; i < n; ++i)

#define sz(a) (int)a.size()

#define lson l , m , rt << 1

#define rson m + 1 , r , rt << 1 | 1

#define bitCount(a)  __builtin_popcount(a)

template<class T> int gmax(T &a, T b) { if (b > a) { a = b; return 1; } return 0; }

template<class T> int gmin(T &a, T b) { if (b < a) { a = b; return 1; } return 0; }

using namespace std;

const int INF = 0x3f3f3f3f;

string to_string(string s) { return '"' + s + '"'; }

string to_string(const char* s) { return to_string((string) s); }

string to_string(bool b) { return (b ? "true" : "false"); }

template <typename A, typename B>

string to_string(pair<A, B> p) { return "(" + to_string(p.first) + ", " + to_string(p.second) + ")"; }

template <typename A>

string to_string(A v) { bool first = true; string res = "{"; for (const auto &x : v) { if (!first) { res += ", "; } first = false; res += to_string(x); } res += "}"; return res; }

void debug_out() { cerr << endl; }

template <typename Head, typename... Tail>

void debug_out(Head H, Tail... T) { cerr << " " << to_string(H); debug_out(T...); }

#ifdef LOCAL

#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)

#else

#define debug(...) 42

#endif

struct Node{

    int u, d, l, r;

    int val;

    Node() {

        u = d = l = r = -1; val = 0;

    }

};

vector<Node> mp;

int R, C;

int getId(int x, int y) { return x * (C + 2) + y; }

void erase(int x) {

   mp[mp[x].l].r = mp[x].r;

   mp[mp[x].r].l = mp[x].l;

   mp[mp[x].u].d = mp[x].d;

   mp[mp[x].d].u = mp[x].u;

   mp[x].val = 0;

}

vector<int> update(vector<int> &choosList, ll &ans, ll &origin) {

    // debug(origin);

    vector<int> needErase;

    vector<int> newList, _newList;

    ans += origin;

    for(int i = 0, len = choosList.size(); i < len; ++i) {

        int x = choosList[i];

        int neiNum = 0; int neiVal = 0;

        if(mp[mp[x].r].val != 0) { neiNum ++; neiVal += mp[mp[x].r].val; }

        if(mp[mp[x].l].val != 0) { neiNum ++; neiVal += mp[mp[x].l].val; }

        if(mp[mp[x].u].val != 0) { neiNum ++; neiVal += mp[mp[x].u].val; }

        if(mp[mp[x].d].val != 0) { neiNum ++; neiVal += mp[mp[x].d].val; }

        // debug(x / (C + 2), x % (C + 2), neiVal, neiNum, mp[x].val);

        if(neiVal > mp[x].val * neiNum) {

            // debug("erase", x / (C + 2), x % (C + 2));

            origin -= mp[x].val;

            needErase.push_back(x);

        }

    }

    for(auto x : needErase) {

        assert(mp[x].r != -1); assert(mp[x].l != -1); assert(mp[x].u != -1); assert(mp[x].d != -1);

        if(mp[mp[x].r].val != 0) { _newList.push_back(mp[x].r); }

        if(mp[mp[x].l].val != 0) { _newList.push_back(mp[x].l); }

        if(mp[mp[x].u].val != 0) { _newList.push_back(mp[x].u); }

        if(mp[mp[x].d].val != 0) { _newList.push_back(mp[x].d); }

        erase(x);

    }

    for(auto it : _newList) {

        if(mp[it].val != 0) newList.push_back(it);

    }

    sort(newList.begin(), newList.end());

    newList.erase(unique(newList.begin(), newList.end()), newList.end());

    // for(int i = 0, len = newList.size(); i < len; ++i) printf("%d %d: ", newList[i] / (C + 2), newList[i] % (C + 2)); printf("\n");

    return newList;

}

int main() {

    int T;

    scanf("%d", &T);

    for(int cas = 1; cas <= T; ++cas) {

        mp.clear();

        scanf("%d %d", &R, &C);

        mp.resize( (R + 5) * (C + 5), Node());

        ll origin = 0;

        for(int i = 1; i <= R; ++i) {

            for(int j = 1; j <= C; ++j) {

                scanf("%d", &mp[getId(i , j)].val);

                origin += mp[getId(i , j)].val;

            }

        }

        for(int i = 1; i <= R; ++i) {

            mp[getId(i , 1)].l = getId(i, 0);

            for(int j = 1; j <= C; ++j) {

                mp[getId(i , j - 1)].r = getId(i, j);

                mp[getId(i , j + 1)].l = getId(i, j);

            }

            mp[getId(i , C)].r = getId(i, C + 1);

        }

        for(int i = 1; i <= C; ++i) {

            mp[getId(1 , i)].u = getId(0, i);

            for(int j = 1; j <= R; ++j) {

                mp[getId(j - 1, i)].d = getId(j, i);

                mp[getId(j + 1, i)].u = getId(j, i);

            }

            mp[getId(R, i)].d = getId(R + 1, i);

        }

        vector<int> choosList;

        for(int i = 1; i <= R; ++i) {

            for(int j = 1; j <= C; ++j) {

                choosList.push_back(getId(i, j));

            }

        }

        ll ans = 0;

        while(1) {

            choosList = update(choosList, ans, origin);

            if(choosList.size() == 0) break;

        }

        printf("Case #%d: %lld\n", cas, ans);

    }

    return 0;

}

/*

4

1 1

15

3 3

1 1 1

1 2 1

1 1 1

1 3

3 1 2

1 3

1 2 3

3 3

1 100 1

1 2 2

1000 1 1

1 3

1 1

*/

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