-图论-最短路- [洛谷 P2243]电路维修

题目描述

Link

Solution

打死也不用纯搜索

这题我们可以考虑最短路

如果一个字符是\,那么我们可以从左上到右下连一条权值为零的边,表示从左上到右下不需要旋转,从右上到左下连一条权值为1的边,表示需要旋转;如果是/,那么则相反

Code

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#define MAXN 1001000
struct Edge {
int v, nx, w;
}e[MAXN << 2];
int head[MAXN], ecnt, n, m, x, y, dis[MAXN], T;
bool vis[MAXN];
char ele[MAXN];
void add(int f, int t, int w) {
e[++ecnt] = (Edge) {t, head[f], w};
head[f] = ecnt;
}
struct node {
int id, w;
};
bool operator < (node a, node b) {
return a.w > b.w;
}
void dijkstra(int s) {
memset(dis, 0x7f, sizeof(dis));
memset(vis, 0, sizeof(vis));
std::priority_queue <node> q;
dis[s] = 0;
q.push((node) {s, 0});
while (!q.empty()) {
node v = q.top();
q.pop();
int u = v.id;
if (vis[u]) continue;
vis[u] = 1;
for (int i = head[u]; i; i = e[i].nx) {
int to = e[i].v;
if (dis[to] > dis[u] + e[i].w) {
dis[to] = dis[u] + e[i].w;
q.push((node) {to, dis[to]});
}
}
}
}
int g(int i, int j) {
return (i - 1) * (m + 1) + j;
}
int main() {
scanf("%d", &T);
while (T--) {
memset(head, 0, sizeof(head));
memset(dis, 0x7f, sizeof(dis));
ecnt = 0;
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) {
scanf("%s", ele + 1);
for (int j = 1; j <= m; j++) {
if (ele[j] == '/') {
add(g(i, j), g(i + 1, j + 1), 1);
add(g(i + 1, j + 1), g(i, j), 1);
add(g(i + 1, j), g(i, j + 1), 0);
add(g(i, j + 1), g(i + 1, j), 0);

}
else {
add(g(i, j), g(i + 1, j + 1), 0);
add(g(i + 1, j + 1), g(i, j), 0);
add(g(i + 1, j), g(i, j + 1), 1);
add(g(i, j + 1), g(i + 1, j), 1);
}
}
}
dijkstra(g(1, 1));
if (dis[g(n + 1, m + 1)] > 2000000000) {
puts("NO SOLUTION");
continue;
}
printf("%d\n", dis[g(n + 1, m + 1)]);
}
}
文章作者: RiverFun
文章链接: https://stevebraveman.github.io/blog/2019/03/13/69/
版权声明: 本博客所有文章除特别声明外,均采用 CC BY-NC-SA 4.0 许可协议。转载请注明来自 RiverFun

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