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;
}