112 lines
3.8 KiB
C++
112 lines
3.8 KiB
C++
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#include <bits/stdc++.h>
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using namespace std;
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typedef pair<int, int> PII;
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/*
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举栗子： 以3行5列为例，共3*5=15个格子
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是4行*6列=24个点
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推广公式化:(n+1)*(m+1)
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*/
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const int N = 510 * 510;
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const int M = 4 * N; // 1个格子里面有4条边，分别为 a->b,b->a,c->d,d->c
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const int INF = 0X3f3f3f3f;
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int n, m;
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int h[N], e[M], w[M], ne[M], idx;
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int dist[N];
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bool st[N];
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void add(int a, int b, int c) {
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e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
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}
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int getNum(int x, int y) {
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/*
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* 按点看 ，为 0,1,2,3,4,...,m
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* 按格子看，为 1,2,3,4,...,m
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即按点看，每行m+1个点
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推广公式化： 编号 = 行号(可以从0开始，也可以从1开始)*列的数量 + 列号
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*/
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return x * (m + 1) + y; //按点看，共m+1列
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}
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int dijkstra() {
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//初始化最短距离
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memset(dist, 0x3f, sizeof dist);
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dist[0] = 0; // 0号点的最短距离是0
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// greater<>()小顶层堆
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priority_queue<PII, vector<PII>, greater<PII>> q;
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q.emplace(0, 0); //注意:先是距离，后是序号！之所以这样设计，是因为小顶堆的排序是默认安排PII的第一维x进行由小到大
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while (q.size()) {
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auto t = q.top();
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q.pop();
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int u = t.second, distance = t.first;
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//如果已经走过,不需要再尝试
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if (st[u]) continue;
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//标识走过
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st[u] = true;
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//枚举每个出边
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i];
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if (dist[j] > distance + w[i]) { //在原距离distanc基础上，再走w[i]后可以到达j,如果这样走近，则更新dist[j]
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dist[j] = distance + w[i];
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q.emplace(dist[j], j); //将j加入队列，继续扩展
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}
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}
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}
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//计算最后一个坐标位置对应的节点，它离起点的最短距离是多少
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return dist[getNum(n, m)];
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}
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signed main() {
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//加快读入
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cin.tie(0), ios::sync_with_stdio(false);
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int T;
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cin >> T;
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while (T--) {
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//还原是否走过的状态st=false
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memset(st, false, sizeof st);
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//多组测试数据，清空邻接表
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memset(h, -1, sizeof h);
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idx = 0;
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// n行m列
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cin >> n >> m;
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//读入数据+建图
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for (int i = 1; i <= n; i++) {
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for (int j = 1; j <= m; j++) {
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char op;
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cin >> op;
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//现在读入的是当前(i,j)坐标的格子中操作符
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/*
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点坐标(0,0) 点坐标(0,1) 点坐标(i-1,j-1) 点坐标(i-1,j)
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格子坐标：(1,1) => 格子坐标：(i,j)
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点坐标(1,0) 点坐标(1,1) 点坐标(i,j-1) 点坐标(i,j)
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*/
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int a = getNum(i - 1, j - 1);
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int b = getNum(i, j);
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int c = getNum(i, j - 1);
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int d = getNum(i - 1, j);
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// Q:为什么权值定义的是 /:1 \:0 呢？
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// A:因为我们描述的是 a->b 的边:
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// 如果a->b之间有边，就是左上角到右下角有边, \ , 权值为0.
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// 如果a->b之间无边，就是右上角到左下角有边, / , 此时需要把它转过来，权值为1.
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int w = op == '/' ? 1 : 0;
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add(a, b, w), add(b, a, w); //无向图，双向建边
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add(c, d, !w), add(d, c, !w); //取反也建边，模拟01边权的无向图
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}
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}
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//求最短路
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int res = dijkstra(); // dijkstra可以处理非负权边的图
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if (res == INF)
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puts("NO SOLUTION");
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else
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printf("%d\n", res);
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}
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return 0;
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}
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