HDU 4856 tunnels (BFS + pressure DP)

HDU 4856 tunnels

Question Link

Given some pipelines, there is no time between pipelines. There are obstacles on the land. It takes 1 time to traverse all pipelines, each pipe can only go once

Idea: first, BFs pre-processes the distance between two pipelines, and then compresses the state to solve the problem. DP [s] [I] indicates the State S, and the minimum cost when stopping the pipeline I

Code:

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;const int INF = 0x3f3f3f3f;const int N = 20;const int d[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};typedef pair<int, int> pii;#define MP(a,b) make_pair(a,b)int g[N][N], vis[N][N], n, m, dp[(1<<15)][20];char G[N][N];struct Pipe {    int x1, y1, x2, y2;    void read() {scanf("%d%d%d%d", &x1, &y1, &x2, &y2);    }} p[N];int bfs(Pipe a, Pipe b) {    queue<pii> Q;    memset(vis, -1, sizeof(vis));    Q.push(MP(a.x2, a.y2));    vis[a.x2][a.y2] = 0;    while (!Q.empty()) {pii now = Q.front();if (now.first == b.x1 && now.second == b.y1) return vis[now.first][now.second];Q.pop();for (int i = 0; i < 4; i++) {    int xx = now.first + d[i][0];    int yy = now.second + d[i][1];    if (xx <= 0 || xx > n || yy <= 0 || yy > n || vis[xx][yy] != -1 || G[xx][yy] != '.') continue;    vis[xx][yy] = vis[now.first][now.second] + 1;    Q.push(MP(xx,yy));}    }    return -1;}void build() {    for (int i = 1; i <= m; i++) {for (int j = 1; j <= m; j++) {    if (i == j) g[i][j] = 0;    else g[i][j] = bfs(p[i], p[j]);}    }}int solve() {    memset(dp, INF, sizeof(dp));    for (int i = 1; i <= m; i++)dp[1<<(i - 1)][i] = 0;    int ans = INF;    for (int i = 0; i < (1<<m); i++) {for (int j = 1; j <= m; j++) {    if (i&(1<<(j - 1))) {for (int k = 1; k <= m; k++) {    if (i&(1<<(k - 1)) == 0 || g[k][j] == -1) continue;    dp[i][j] = min(dp[i^(1<<(j - 1))][k] + g[k][j], dp[i][j]);}    }    if (i == (1<<m) - 1)ans = min(ans, dp[i][j]);}    }    if (ans == INF) return -1;    return ans;}int main() {    while (~scanf("%d%d", &n, &m)) {for (int i = 1; i <= n; i++)    scanf("%s", G[i] + 1);for (int i = 1; i <= m; i++)    p[i].read();build();printf("%d\n", solve());    }    return 0;}