acwing-175. 电路维修

考点

  • 双端队列BFS

题解

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#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e2 + 50;
int r, c, d[maxn][maxn];
char s[maxn][maxn];
vector<pair<pair<int, int>, int>> p[maxn][maxn];
bool v[maxn][maxn];
deque<pair<int, int>> q;

void add(int x1, int y1, int x2, int y2, int z) {
p[x1][y1].push_back(make_pair(make_pair(x2, y2), z));
}

void init() {
memset(v, 0, sizeof(v));
for (int i = 0; i <= r; ++i) {
for (int j = 0; j <= c; ++j) {
p[i][j].clear();
}
}
q.clear();
for (int i = 1; i <= r; ++i) {
for (int j = 1; j <= c; ++j) {
if (s[i][j] == '/') {
add(i - 1, j - 1, i, j, 1);
add(i, j, i - 1, j - 1, 1);
add(i, j - 1, i - 1, j, 0);
add(i - 1, j, i, j - 1, 0);
} else {
add(i - 1, j - 1, i, j, 0);
add(i, j, i - 1, j - 1, 0);
add(i, j - 1, i - 1, j, 1);
add(i - 1, j, i, j - 1, 1);
}
}
}
memset(d, 0x3f, sizeof(d));
}

void dijkstra() {
d[0][0] = 0;
q.push_back(make_pair(0, 0));
while (!q.empty()) {
int x = q.front().first, y = q.front().second;
q.pop_front();
v[x][y] = 1;
if (x == r && y == c) {
cout << d[r][c] << endl;
return;
}
for (auto it : p[x][y]) {
int nx = it.first.first, ny = it.first.second, nz = it.second;
if (v[nx][ny]) continue;
if (d[x][y] + nz < d[nx][ny]) {
d[nx][ny] = d[x][y] + nz;
if (nz) {
q.push_back(make_pair(nx, ny));
} else {
q.push_front(make_pair(nx, ny));
}
}
}
}
}

int main() {
int n;
cin >> n;
while (n--) {
cin >> r >> c;
for (int i = 1; i <= r; ++i) cin >> (s[i] + 1);
if ((r & 1) != (c & 1)) {
cout << "NO SOLUTION" << endl;
continue;
}
init(), dijkstra();
}
return 0;
}

思路

双端队列BFS的模板题,直接上图。