SCOI2009 最长距离

洛谷

BZOJ

分析

图论题。直接 \(dfs\) 加一点剪枝就能过了。

代码

#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;
}