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UVA 1078 - Steam Roller(最短路)

来源:程序员人生   发布时间:2014-10-10 08:00:00 阅读次数:3300次

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 "); else printf("%d ", ans); } return 0; }


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