Google Code Jam 2020 Round 1A 2020 题解

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
*/
posted @ 2020-04-12 12:15  basasuya  阅读(435)  评论(0编辑  收藏  举报