描述
题解
这个题真的有趣,最短路,难点主要是在建图上~~~
这里先说一下题面的误区,我一开始理解为每盏灯只能点亮其所在的行或者列,谁成想并不是这样,而是说能点亮任何行或者列。那是否意味着我们需要考虑他和任何一行或者一列的边呢?实际上并不需要,因为题目中有一个很强的条件,当他要点亮一行或者一列时,前边点亮的就会熄灭,而人无时不刻都要待在光亮区域,所以呢?在每次点亮时,人都要在有灯的地方才行,酱紫的话,我们经过思考后,其实每次点亮时,都只有六个部分有价值,除去灯所在的行或者列外,剩下的四个就是灯所在的行或者列相邻的行或者列,这里我们将这些行或者列抽象为一个点,建边,权值为 1 ,反向建边,权值为
最短路好题,建图套路深!
代码
#include <iostream>
#include <cstring>
#include <vector>
#include <queue>
#include <map>
using namespace std;
typedef pair<int, int> pii;
const int MAXN = 1e5;
const int OFFSET = 1e4 + 10;
const int INF = 0x3f3f3f3f;
const int DIR[4][2] = {
{
0, 1}, {
0, -1},
{
1, 0}, {-1, 0}};
int cnt = 1;
struct edge
{
int to, cost;
edge() {}
edge(int to_, int cost_) : to(to_), cost(cost_) {}
};
void add_edge(vector<edge> G[], int r, int c)
{
G[cnt].push_back(edge(r + OFFSET, 1));
G[cnt].push_back(edge(r + 1 + OFFSET, 1));
G[cnt].push_back(edge(r - 1 + OFFSET, 1));
G[cnt].push_back(edge(c + 2 * OFFSET, 1));
G[cnt].push_back(edge(c + 1 + 2 * OFFSET, 1));
G[cnt].push_back(edge(c - 1 + 2 * OFFSET, 1));
G[r + OFFSET].push_back(edge(cnt, 0));
G[r + 1 + OFFSET].push_back(edge(cnt, 0));
G[r - 1 + OFFSET].push_back(edge(cnt, 0));
G[c + 2 * OFFSET].push_back(edge(cnt, 0));
G[c + 1 + 2 * OFFSET].push_back(edge(cnt, 0));
G[c - 1 + 2 * OFFSET].push_back(edge(cnt, 0));
cnt++;
}
int n, m, k;
int dis[MAXN];
int vis[MAXN];
pii p[MAXN];
vector<edge> E[MAXN];
map<pii, int> mpi;
void dijkstra(int s)
{
memset(dis, 0x3f, sizeof(dis));
memset(vis, 0, sizeof(vis));
priority_queue<pii, vector<pii>, greater<pii> > pq;
dis[s] = 0;
pq.push(pii(0, s));
pii tmp;
while (!pq.empty())
{
tmp = pq.top();
pq.pop();
int u = tmp.second;
if (vis[u])
{
continue;
}
vis[u] = 1;
for (int i = 0; i < E[u].size(); i++)
{
edge e = E[u][i];
if (dis[e.to] > dis[u] + e.cost)
{
dis[e.to] = dis[u] + e.cost;
pq.push(pii(dis[e.to], e.to));
}
}
}
}
int main(int argc, const char * argv[])
{
cin >> n >> m >> k;
int r, c;
for (int i = 1; i <= k; i++)
{
scanf("%d%d", &r, &c);
add_edge(E, r, c);
p[i] = make_pair(r, c);
mpi[p[i]] = i;
}
pii t;
for (int i = 1; i <= k; i++)
{
for (int j = 0; j < 4; j++)
{
int x = p[i].first + DIR[j][0];
int y = p[i].second + DIR[j][1];
t = make_pair(x, y);
if (mpi[t])
{
E[i].push_back(edge(mpi[t], 0));
}
}
}
t = make_pair(n, m);
if (!mpi[t])
{
E[n + OFFSET].push_back(edge(k + 1, 0));
E[m + 2 * OFFSET].push_back(edge(k + 1, 0));
}
dijkstra(mpi[pii(1, 1)]);
int ans;
if (mpi[t])
{
ans = dis[mpi[t]];
}
else
{
ans = dis[k + 1];
}
if (ans == INF)
{
ans = -1;
}
cout << ans << '\n';
return 0;
}