Codeforces373D-Lakes in Berland（dfs）

最新推荐文章于 2018-11-23 12:30:31 发布

Lzed

最新推荐文章于 2018-11-23 12:30:31 发布

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分类专栏： graph codeforces 文章标签： codeforces

本文链接：https://blog.youkuaiyun.com/Lzedo/article/details/52733921

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本文提供了一道Codeforces D级难度题目（题目链接：http://codeforces.com/contest/723/problem/D）的解题思路及代码实现。通过DFS算法处理连通分量，并排除边界上的点，最终统计并输出结果。

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题目链接

http://codeforces.com/contest/723/problem/D

思路

先对格点图跑一遍dfs，对连通分量编个号，然后把边界的点排除掉，最后统计一下就好

代码

#include <iostream>
#include <cstring>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#include <queue>
#include <sstream>
#include <iomanip>
#include <fstream>
#include <cstdio>
#include <cstdlib>
#include <climits>
#include <deque>
#include <bitset>
#include <algorithm>
using namespace std;

#define PI acos(-1.0)
#define LL long long
#define PII pair<int, int>
#define PLL pair<LL, LL>
#define mp make_pair
#define IN freopen("in.txt", "r", stdin)
#define OUT freopen("out.txt", "wb", stdout)
#define scan(x) scanf("%d", &x)
#define scan2(x, y) scanf("%d%d", &x, &y)
#define scan3(x, y, z) scanf("%d%d%d", &x, &y, &z)
#define sqr(x) (x) * (x)

const int maxn = 55;
int vis[maxn][maxn];
char a[maxn][maxn];
int use[5000], n, m, k;
int dx[] = {
    -1, 1, 0, 0
};
int dy[] = {
    0, 0, -1, 1
};

void dfs(int x, int y, int id) {
    vis[x][y] = id;
    for (int i = 0; i < 4; i++) {
        int tx = x + dx[i], ty = y + dy[i];
        if (vis[tx][ty] || tx < 0 || tx >= n || ty < 0 || ty >= m || a[tx][ty] == '*') continue;
        dfs(tx, ty, id);
    }
}

int main() {
    scan3(n, m, k);
    int id = 0;
    for (int i = 0; i < n; i++) scanf("%s", a[i]);
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (!vis[i][j] && a[i][j] == '.') dfs(i, j, ++id);
        }
    }
    for (int j = 0; j < m; j++) use[vis[0][j]] = use[vis[n - 1][j]] = -1;
    for (int i = 0; i < n; i++) use[vis[i][0]] = use[vis[i][m - 1]] = -1;
    int num[5000];
    memset(num, 0, sizeof(num));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            int id = vis[i][j];
            if (use[id] == -1) continue;
            num[id]++;
        }
    }
    priority_queue<PII, vector<PII>, greater<PII> > Q;
    int tot = 0;
    for (int i = 1; i <= id; i++) {
        if (use[i] == -1) continue;
        Q.push(mp(num[i], i));
        tot++;
    }
    int res = 0, t = tot - k;
    while (t) {
        if (Q.empty()) break;
        PII x = Q.top(); Q.pop();
        t--;
        res += x.first;
        use[x.second] = 1;
    }
    cout << res << endl;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (a[i][j] == '*') printf("*");
            else {
                if (use[vis[i][j]] == 1) printf("*");
                else printf(".");
            }
        }
        cout << endl;
    }
    return 0;
}