UVA 1078 - Steam Roller

题目链接

题意：给定一个地图，要求起点走到终点需要的时间，如果进入一个转弯位置，则进入和出去的时候时间要加倍

思路：最短路，关键在于如何建模，每个结点d[x][y][d][flag]表示在x, y结点，方向为d，是否加倍过了，这样就可以把每个结点之间对应的关系建边，做最短路即可

代码：

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;

const int MAXNODE = 100005;
const int MAXEDGE = 1000005;

typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge {
	int u, v;
	Type dist;
	Edge() {}
	Edge(int u, int v, Type dist) {
		this->u = u;
		this->v = v;
		this->dist = dist;
	}
};

struct HeapNode {
	Type d;
	int u;
	HeapNode() {}
	HeapNode(Type d, int u) {
		this->d = d;
		this->u = u;
	}
	bool operator < (const HeapNode& c) const {
		return d > c.d;
	}
};

struct Dijkstra {
	int n, m;
	Edge edges[MAXEDGE];
	int first[MAXNODE];
	int next[MAXEDGE];
	bool done[MAXNODE];
	Type d[MAXNODE];
	int p[MAXNODE];

	void init(int n) {
		this->n = n;
		memset(first, -1, sizeof(first));
		m = 0;
	}

	void add_Edge(int u, int v, Type dist) {
		edges[m] = Edge(u, v, dist);
		next[m] = first[u];
		first[u] = m++;
	}

	void dijkstra(int s) {
		priority_queue<HeapNode> Q;
		for (int i = 0; i < n; i++) d[i] = INF;
		d[s] = 0;
		p[s] = -1;
		memset(done, false, sizeof(done));
		Q.push(HeapNode(0, s));
		while (!Q.empty()) {
			HeapNode x = Q.top(); Q.pop();
			int u = x.u;
			if (done[u]) continue;
			done[u] = true;
			for (int i = first[u]; i != -1; i = next[i]) {
				Edge& e = edges[i];
				if (d[e.v] > d[u] + e.dist) {
					d[e.v] = d[u] + e.dist;
					p[e.v] = i;
					Q.push(HeapNode(d[e.v], e.v));
				}
			}
		}
	}
} gao;

// 0 up, 1 left, 2 right, 3 down
const int D[4][2] = {-1, 0, 0, -1, 0, 1, 1, 0};

int read_init() {
	int x; scanf("%d", &x);
	return x;
}

const int N = 105;
int R, C, r1, c1, r2, c2;
int grid[N][N][4], id[N][N][4][2];

int get(int r, int c, int d, int flag) {
	if (id[r][c][d][flag] == -1) id[r][c][d][flag] = gao.n++;
	return id[r][c][d][flag];
}

bool can(int r, int c, int d) {
	if (r < 0 || r >= R || c < 0 || c >= C) return false;
	return grid[r][c][d] > 0;
}

int main() {
	int cas = 0;
	while (~scanf("%d%d%d%d%d%d", &R, &C, &r1, &c1, &r2, &c2) && R) {
		r1--; c1--; r2--; c2--;
		memset(id, -1, sizeof(id));
		memset(grid, 0, sizeof(grid));
		gao.init(1);
		for (int i = 0; i < R; i++) {
			for (int j = 0; j < C - 1; j++)
				grid[i][j + 1][1] = grid[i][j][2] = read_init();
			if (i == R - 1) continue;
			for (int j = 0; j < C; j++)
				grid[i][j][3] = grid[i + 1][j][0] = read_init();
		}
		for (int i = 0; i < 4; i++) {
			if (!can(r1, c1, i)) continue;
			gao.add_Edge(0, get(r1 + D[i][0], c1 + D[i][1], i, 1), grid[r1][c1][i] * 2);
		}
		for (int i = 0; i < R; i++) {
			for (int j = 0; j < C; j++) {
				for (int d1 = 0; d1 < 4; d1++) {
					if (!can(i, j, 3 - d1)) continue;
					for (int d2 = 0; d2 < 4; d2++) {
						if (!can(i, j, d2)) continue;
						for (int flag = 0; flag < 2; flag++) {
							int x = i + D[d2][0];
							int y = j + D[d2][1];
							int v = grid[i][j][d2];
							int ff = 0;
							if (d1 != d2) {
								if (!flag) v += grid[i][j][3 - d1];
								ff = 1; v += grid[i][j][d2];
							}
							gao.add_Edge(get(i, j, d1, flag), get(x, y, d2, ff), v);
						}
					}
				}
			}
		}
		gao.dijkstra(0);
		int ans = INF;
		for (int i = 0; i < 4; i++) {
			if (!can(r2, c2, 3 - i)) continue;
			for (int j = 0; j < 2; j++) {
				int v = gao.d[get(r2, c2, i ,j)];
				if (!j) v += grid[r2][c2][3 - i];
				ans = min(v, ans);
			}
		}
		printf("Case %d: ", ++cas);
		if (ans == INF) printf("Impossible\n");
		else printf("%d\n", ans);
	}
	return 0;
}