codeforce round #217

A题：

rook的话如果同行同列就一次，否则两次。

bishop，首先如果一个在白色位置，另外一个在黑色位置是不可达的。

否则再一条对角线上就一次，其余位置两次。

king，bfs()一次求出答案。

AC代码：

#include <cstdio>
#include <cstring>


const int MAX_NUMBER = 1007;

struct Point {
    int x, y, steps;
};
int n;
int start_x, start_y, end_x, end_y;
int ans_1, ans_2, ans_3;
int step[8][2] = {{-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 1}, {1, -1}, {1, 0}, {1, 1}};
int vis[10][10];
struct Point point[MAX_NUMBER];

void bfs() {
    memset(vis, 0, sizeof(vis));
    point[0].x = start_x;
    point[0].y = start_y;
    vis[start_x][start_y] = 1;
    int front = 0;
    int rear = 1;
    while (front < rear) {
        struct Point cnt_point = point[front];
        int cnt_x = cnt_point.x;
        int cnt_y = cnt_point.y;
        if (cnt_x == end_x && cnt_y == end_y) {
            ans_3 = cnt_point.steps;
            return ;
        }
        for (int i = 0; i < 8; i++) {
            int next_x = cnt_x + step[i][0];
            int next_y = cnt_y + step[i][1];
            if (next_x >= 1 && next_x <= 8 && next_y >= 1 && next_y <= 8 && !vis[next_x][next_y]) {
                vis[next_x][next_y] = 1;
                point[rear].x = next_x;
                point[rear].y = next_y;
                point[rear].steps = cnt_point.steps + 1;
                rear++;
            }
        }
        front++;
    }
}
int main() {

    scanf("%d%d%d%d", &start_x, &start_y, &end_x, &end_y);
    if (start_x == end_x || start_y == end_y) {
        ans_1 = 1;
    }
    else {
        ans_1 = 2;
    }
    if ((start_x + start_y) % 2 != (end_x + end_y) % 2) {
        ans_2 = 0;
    }
    else if (start_x + start_y == end_x + end_y || start_x - start_y == end_x - end_y) {
        ans_2 = 1;
    }
    else {
        ans_2 = 2;
    }
    bfs();
    printf("%d %d %d\n", ans_1, ans_2, ans_3);
    return 0;
}

B题：

O(n ^ 2)，枚举判断一个集合是不是另一个集合的子集或者真子集。

如果是真子集的话，那么超集就不可能赢。如果是想等的话两个都不可能赢，其他情况可能赢。

AC代码：

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <set>

using namespace std;
const int MAX_NUMBER = 107;
vector<int> player[MAX_NUMBER];
set<int> player_set[MAX_NUMBER];
int vis[MAX_NUMBER], ans[MAX_NUMBER];
int player_number;
int main() {

    memset(vis, 0, sizeof(vis));
    scanf("%d", &player_number);
    for (int i = 1; i <= player_number; i++) {
        int cnt_number;
        scanf("%d", &cnt_number);
        for (int j = 1; j <= cnt_number; j++) {
            int temp;
            scanf("%d", &temp);            
            player_set[i].insert(temp);
            player[i].push_back(temp);
        }
    }
    int n = 1;
    memset(ans, -1, sizeof(ans));
    for (int i = 1; i <= player_number; i++) {
        for (int j = i + 1; j <= player_number; j++) {
            int flag = 0;
            if (player[i].size() == player[j].size()) {
                for (int k = 0; k < player[j].size(); k++) {
                    if (!player_set[i].count(player[j][k])) {
                        flag = 1;
                        break;
                    }
                }
                if (!flag) {
                    ans[i] = ans[j] = 0;
                }
            }
            else if (player[i].size() < player[j].size()) {
                for (int k = 0; k < player[i].size(); k++) {
                    if (!player_set[j].count(player[i][k])) {
                        flag = 1;
                        break;
                    }
                }
                if (!flag) {
                    ans[j] = 0;
                }
            }
            else {
                for (int k = 0; k < player[j].size(); k++) {
                    if (!player_set[i].count(player[j][k])) {
                        flag = 1;
                        break;
                    }
                }
                if (!flag) {
                    ans[i] = 0;
                }
            }
        }
    }
    for (int i = 1; i <= player_number; i++) {
        if (ans[i] == -1) {
            printf("YES\n");
        }
        else {
            printf("NO\n");
        }
    }
    return 0;
}

C题：

贪心。

每次对各种颜色有的手套数进行排序，颜色最少的和颜色最多的去匹配。

匹配到最后无法匹配的只能和自己匹配了。

但是这种贪心不是正确的，比如说颜色1：7个人，颜色2：5个人，颜色3：1个人，颜色4：3个人。

按这种贪心的话，

先匹配3- 1，1-3，然后4-1，1-4，4-1，1-4，接下来再1-2，2-1，1-2，2-1，1-2，2-1

最后颜色2还剩下4只手套，只能是2-2，2-2，但是有一种明显更好的方式，比如说最后的2和3-1的随便一个进行交换，答案会多出来。

所有我的做法是先贪心，贪心完后看是否能不能和前面的手套进行交换使结果更好，如果能的话肯定要交换，得出最后的答案。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int MAX_NUMBER = 100007;

struct Color {
    int color, number;
};
struct Ans {
    int x, y;
};

struct Ans ans[MAX_NUMBER];
bool cmp(struct Color a, struct Color b) {
    return a.number < b.number;
}
struct Color color[MAX_NUMBER];
int vis[MAX_NUMBER];
int color_number, ans_number, children, pre_number, diff;

void deal() {
    pre_number = color_number;
    sort(color + 1, color + color_number + 1, cmp);
    color_number = 0;
    for (int i = 1; i <= pre_number; i++) {
        if (color[i].number != 0) {
            color[++color_number].color = color[i].color;
            color[color_number].number = color[i].number;
        }
    }
    if (color_number >= 2) {
        for (int i = 0; i < color[1].number; i += 2) {
            ans[ans_number].y = color[1].color;
            ans[ans_number].x = color[color_number].color;
            ans_number++;
            ans[ans_number].x = color[1].color;
            ans[ans_number].y = color[color_number].color;
            ans_number++;
        }
        color[color_number].number -= color[1].number;     
        color[1].number = 0;
        diff = ans_number;
    }
    else if (color_number == 1){
        for (int i = 0; i < color[1].number; i += 2) {
            ans[ans_number].x = color[1].color;
            ans[ans_number].y = color[1].color;
            ans_number++;
        }
        color_number = 0;
    }
}
int main() {

    int m;
    scanf("%d%d", &children, &m);
    memset(color, 0, sizeof(color));
    memset(vis, 0, sizeof(vis));
    color_number = 0;
    for (int i = 1; i <= m; i++) {
        color[i].color = i;
    }
    for (int i = 1; i <= children; i++) {
        int temp;
        scanf("%d", &temp);
        if (!vis[temp]) {
            vis[temp] = 1;
            color[temp].number = 2;
        }
        else {
            color[temp].number += 2;
        }
    }
    color_number = m;
    while (1) {
        deal();
        if (color_number == 0) {
            break;
        }
    }
    int final_result = diff;    
    if (ans_number != diff) {
        int cnt_number = 0;
        for (int i = 0; i < diff && cnt_number < ans_number - diff; i++) {
            if (ans[i].x != ans[diff + cnt_number].x && ans[i].y != ans[diff + cnt_number].y) {
                int temp = ans[diff + cnt_number].x;
                ans[diff + cnt_number].x = ans[i].x;
                ans[i].x = temp;
                final_result++;
                cnt_number++;
            }
        }
    }
    printf("%d\n", final_result);
    for (int i = 0; i < ans_number; i++) {
        printf("%d %d\n", ans[i].x, ans[i].y);
    }
    return 0;
}

D题：

没看懂题目意思= =