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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)
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