洛谷

BZOJ

分析

圖論題。直接 \(dfs\) 加一點(diǎn)剪枝就能過了。

代碼

#include <cmath>
#include <cstdio>
#include <iomanip>
#include <cstring>
#include <iostream>
#include <algorithm>
#define il inline
#define re register
#define tie0 cin.tie(0),cout.tie(0)
#define fastio ios::sync_with_stdio(false)
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
typedef long long ll;template <typename T> inline void read(T &x) {T f = 1; x = 0; char c;for (c = getchar(); !isdigit(c); c = getchar()) if (c == '-') f = -1;for ( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);x *= f;
}const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, -1, 0, 1};
int n, m, t, ans;
int g[35][35], dis[35][35];void dfs(int x, int y, int step) {if (step > t) return;if (dis[x][y] <= step) return;dis[x][y] = step;for (int i = 0; i < 4; ++i) {int nx = x + dx[i], ny = y + dy[i];if (nx <= n && nx > 0 && ny <= m && ny >0)dfs(nx, ny, step + g[nx][ny]);}
}int main() {char ch;read(n), read(m), read(t);for (int i = 1; i <= n; ++i)for (int j = 1; j <= m; ++j) {cin >> ch;g[i][j] = ch - '0';}for (int i = 1; i <= n; ++i)for (int j = 1; j <= m; ++j) {memset(dis, 127, sizeof(dis));dfs(i, j, g[i][j]);for (int k = 1; k <= n; ++k)for (int l = 1; l <= m; ++l)if (dis[k][l] <= t)ans = max(ans, (i-k) * (i-k) + (j-l) * (j-l));}cout << fixed << setprecision(6) << sqrt(ans);return 0;
}

轉(zhuǎn)載于:https://www.cnblogs.com/hlw1/p/11437215.html

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