TZOJ4521: Flooded Island（简单搜索判断、矩形最小边界最大边界）

原创已于 2023-08-17 00:28:35 修改 · 117 阅读

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于 2023-08-17 00:27:51 首次发布

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文章讲述了如何解决一个关于海平面上升导致岛屿土地被淹没的问题，需要找出包含所有陆地的最小受破坏矩形区域。解题思路是遍历地图，检查每个陆地是否被至少3个海洋方块包围，然后输出更新后的最小边界矩形地图。

TZOJ4521: Flooded Island

题目传送门

描述

There is an beautiful island on the sea, but recently the sea level arround the island is raised, and continuously the land is disrupted. Now, we want to know what it will be like some years later.

The island is describled as a grid of R*C squares, each square has a character:

(1)‘X’: represents the land;

(2)‘.’: represents the sea.
If the land square is currently surrounded by at least 3 sea squares in four direction(north, south, east and west), it will be disrupted.

Now, please output the smallest disrupted map that contains all land squares.

There has at least one square of land will remain in all test cases.

We also assume that the map are surrounded by sea squares.

输入

The input case described the current map of the island.
The first line has two positive integers, R and C (1 <= R, C <= 10), then followed by R lines. Each line has C characters.

输出

Output the rectangluar map of the disrupted island that contains all land squares after some years later.

样例输入

5 3
...
.X.
.X.
.X.
...

样例输出

解题思路

本题主要还是题意理解，原题目中的样例给得不好理解，换成下面就好理解了。原题在洛谷。
3 10
…
…XXX.XXX.
XXX…
输出
.XX…X
XX…
这里就是把周围有3个以上.（超出边界也算.）的X变成.，然后找出矩形的最小边界最大边界输出矩形图即可。

代码

#include<bits/stdc++.h>
using namespace std;
#define ll int
#define inf 0x3f3f3f3f
#define IOS ios::sync_with_stdio(0), cin.tie(0)
const ll N = 1e1 * 1 + 5;
ll n, m;
char a[N][N], b[N][N];
ll dir[][2] = {1, 0, -1, 0, 0, 1, 0, -1};
void solve() {
    cin >> n >> m;
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
    }
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            if(a[i][j] == 'X') {
                ll ct = 0;
                for (int k = 0; k < 4; ++k) {
                    ll x = i + dir[k][0];
                    ll y = j + dir[k][1];
                    if(a[x][y] != 'X') ++ ct;
                }
                if(ct < 3) b[i][j] = 'X';
                else b[i][j] = '.';
            } else {
                b[i][j] = a[i][j];
            }
        }
    }
    ll u = inf, d = 0, l = inf, r = 0;
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            if(b[i][j] == 'X') {
                u = min(u, i);
                d = max(d, i);
                l = min(l, j);
                r = max(r, j);
            }
        }
    }
    for (int i = u; i <= d; ++i) {
        for (int j = l; j <= r; ++j) {
            cout << b[i][j];
        }
        cout << endl;
    }
}

int main( ){
    IOS;
//    ll t; cin >> t;
    ll t = 1;
    while (t--) solve();
    return 0;
}