hdu 5433 Xiao Ming climbing（最短路）

题目链接：hdu 5433 Xiao Ming climbing

三维状态，x，y，k，用优先队列优化的dijkstra做一遍。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <algorithm>

using namespace std;
const int maxn = 55;
const double eps = 1e-6;
const double inf = 1e20;
const int dir[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

int N, M, K, Sx, Sy, Ex, Ey;
bool done[maxn][maxn][maxn];
char G[maxn][maxn];
double dp[maxn][maxn][maxn];

struct State {
	int x, y, k;
	double d;
	State(int x = 0, int y = 0, int k = 0, double d = 0): x(x), y(y), k(k), d(d) {}
	bool operator < (const State& u) const { return d > u.d; }
};

void init () {
	scanf("%d%d%d", &N, &M, &K);
	for (int i = 1; i <= N; i++)
		scanf("%s", G[i]+1);
	scanf("%d%d%d%d", &Sx, &Sy, &Ex, &Ey);
}

bool Dijkstra () {
	for (int i = 1; i <= N; i++)
		for (int j = 1; j <= M; j++)
			for (int k = 0; k <= K; k++) dp[i][j][k] = inf;
	memset(done, false, sizeof(done));

	dp[Sx][Sy][K] = 0;
	priority_queue<State> Q;
	Q.push(State(Sx, Sy, K, 0));

	while (!Q.empty()) {
		State u = Q.top();
		Q.pop();
		int x = u.x, y = u.y, k = u.k;

		if (done[x][y][k]) continue;
		done[x][y][k] = true;

		for (int i = 0; i < 4; i++) {
			int p = x + dir[i][0];
			int q = y + dir[i][1];
			int t = k - 1;
			if (p <= 0 || p > N || q <= 0 || q > M || t <= 0 || G[p][q] == '#') continue;
			double d = 1.0 * abs(G[x][y] - G[p][q]) / k;
			if (dp[p][q][t] > dp[x][y][k] + d) {
				dp[p][q][t] = dp[x][y][k] + d;
				Q.push(State(p, q, t, dp[p][q][t]));
			}
		}
	}

	double ans = inf;
	for (int i = 1; i <= K; i++)
		ans = min(ans, dp[Ex][Ey][i]);
	if (fabs(ans-inf) < eps) return false;

	printf("%.2lf\n", ans);
	return true;
}

int main () {
	int cas;
	scanf("%d", &cas);
	while (cas--) {
		init();
		if (!Dijkstra()) printf("No Answer\n");
	}
	return 0;
}