Fundamental Datastructure

11988 - Broken Keyboard (a.k.a. Beiju Text)

可以用deque来模拟。

#include <iostream>

#include <string>

#include <string.h>

#include <stdio.h>

#include <queue>

using namespace std;



const int MAX = 100005;



char ch[MAX];



int main() {

    while (scanf("%s", ch) != EOF) {

        deque<string> Q;

        string buffer = "";

        int toward = 1;

        int n = strlen(ch);

        

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

            if (ch[i] == '[') {

                if (toward > 0)  Q.push_back(buffer);

                else  Q.push_front(buffer);

                toward = -1;

                buffer = "";

            } else if (ch[i] == ']') {

                if (toward > 0)  Q.push_back(buffer);

                else  Q.push_front(buffer);

                toward = 1;

                buffer = "";

            } else {

                buffer += ch[i];

            }

        }



        if (toward > 0)  Q.push_back(buffer);

        else  Q.push_front(buffer);



        for (int i = 0; i < Q.size(); i++) {

            printf("%s", Q[i].c_str());

        }

        printf("\n");

    }

}

 

10410 - Tree Reconstruction

一点点解析。

#include <cstdio>

#include <vector>

using namespace std;



const int MAXN = 1005;



int n, bfs[MAXN], dfs[MAXN];

int idx[MAXN], root[MAXN];



int main() {

    while (scanf("%d", &n) != EOF) {

        for (int i = 0; i < n; i++)  scanf("%d", &bfs[i]);

        for (int i = 0; i < n; i++)  scanf("%d", &dfs[i]);

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

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

                if (bfs[i] == dfs[j]) {

                    idx[i] = j;

                    break;

                }

            }

        }      

        int p = 1;

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

            root[i] = bfs[0];

        }

        vector<vector<int> > tree(n + 1);

        while (p < n) {

            vector<int> sons;

            sons.push_back(p);

            int max_idx = idx[p];

            int i;

            for (i = p + 1; i < n; i++) {

                if (root[idx[i]] != root[idx[p]] || idx[i] < max_idx) {

                    break;

                }

                max_idx = idx[i];

                sons.push_back(i);

            }

            tree[root[idx[p]]] = sons;

            p = i;

            for (i = sons.size() - 1; i >= 0; i--) {

                int j = idx[sons[i]];

                int r = root[j];

                while (j < n && root[j] == r) {

                    root[j++] = bfs[sons[i]];

                }

            }

        }

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

            printf("%d:", j);

            for (int k = 0; k < tree[j].size(); k++) {

                printf(" %d", bfs[tree[j][k]]);

            }

            printf("\n");

        }

    }

}

 

10895 - Matrix Transpose

行列互换后排序。

#include <cstdio>

#include <vector>

#include <algorithm>

using namespace std;



struct Data {

    int r, c, v;

    bool operator < (const Data& d) const {

        if (r != d.r) {

            return r < d.r;

        } else {

            return c < d.c;

        }

    }

};



int main() {

    int m, n;

    while (scanf("%d%d", &m, &n) != EOF) {

        vector<Data> datas;

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

            int r;

            scanf("%d", &r);

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

                Data d;

                scanf("%d", &d.r);

                d.c = i;

                datas.push_back(d);

            }

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

                scanf("%d", &datas[datas.size() - r + j].v);

            }

        }

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

        int mm = 1, i = 0;

        printf("%d %d\n", n, m);

        while (mm <= n) {

            vector<Data> row;

            while (i < datas.size() && datas[i].r == mm) {

                row.push_back(datas[i]);

                i++;

            }

            mm++;

            printf("%d", (int)row.size());

            for (int j = 0; j < row.size(); j++) {

                printf(" %d", row[j].c);

            }

            printf("\n");

            for (int j = 0; j < row.size(); j++) {

                if (j)  printf(" ");

                printf("%d", row[j].v);

            }

            printf("\n");

        }

    }

}