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CodeForces 723D Lakes in Berland (dfs搜索)

题意:给定一个n*m的矩阵,*表示陆地, . 表示水,一些连通的水且不在边界表示湖,让你填最少的陆地使得图中湖剩下恰好为k。

析:很简单的一个搜索题,搜两次,第一次把每个湖的位置和连通块的数量记下来,第二次去填陆地,选少的进行填。

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>//#include <tr1/unordered_map>#define freopenr freopen("in.txt", "r", stdin)#define freopenw freopen("out.txt", "w", stdout)using namespace std;//using namespace std :: tr1;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const LL LNF = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 2e3 + 5;const LL mod = 10000000000007;const int N = 1e6 + 5;const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){    return r >= 0 && r < n && c >= 0 && c < m;}char s[55][55];struct Node{    int x, y, cnt;    Node(int xx, int yy, int c) : x(xx), y(yy), cnt(c) { }    bool operator < (const Node &p) const{        return cnt < p.cnt;    }};bool vis[55][55];int ans, cnt;void dfs(int r, int c){    ans = Max(ans, cnt);    for(int i = 0; i < 4; ++i){        int x = r + dr[i];        int y = c + dc[i];        if(x < 0 || x >= n || y < 0 || y >= m){ ans = INF;  continue; }        if(vis[x][y] || s[x][y] == ‘*‘)  continue;        vis[x][y] = true;        ++cnt;        dfs(x, y);    }}void dfs1(int r, int c){    for(int i = 0; i < 4; ++i){        int x = r + dr[i];        int y = c + dc[i];        if(!is_in(x, y) || s[x][y] == ‘*‘) continue;        s[x][y] = ‘*‘;        dfs1(x, y);    }}int main(){    int k;    while(scanf("%d %d %d", &n, &m, &k) == 3){        for(int i = 0; i < n; ++i)  scanf("%s", s+i);        vector<Node> v;        memset(vis, false, sizeof vis);        for(int i = 0; i < n; ++i){            for(int j = 0; j < m; ++j){                if(s[i][j] == ‘.‘ && !vis[i][j]){                    ans = 0;                    cnt = 1;                    vis[i][j] = true;                    dfs(i, j);                    if(ans != INF)  v.push_back(Node(i, j, ans));                }            }        }        sort(v.begin(), v.end());        ans = 0;        for(int i = 0; i+k < v.size(); ++i){            s[v[i].x][v[i].y] = ‘*‘;            dfs1(v[i].x, v[i].y);            ans += v[i].cnt;        }        printf("%d\n", ans);        for(int i = 0; i < n; ++i)            puts(s[i]);    }    return 0;}

 

CodeForces 723D Lakes in Berland (dfs搜索)