标签:target std mod code tle [1] air ons memset
题意:就是推箱子游戏,问最少要几步
思路:每一个箱子和目标位置建边。权值为负权值,然后进行二分图完美匹配就可以,注意不能到达的位置权值应该置为最小
代码:
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <queue> using namespace std; const int MAXNODE = 45; typedef int Type; const Type INF = 0x3f3f3f3f; struct KM { int n; Type g[MAXNODE][MAXNODE]; Type Lx[MAXNODE], Ly[MAXNODE], slack[MAXNODE]; int left[MAXNODE]; bool S[MAXNODE], T[MAXNODE]; void init(int n) { this->n = n; } void add_Edge(int u, int v, Type val) { g[u][v] = val; } bool dfs(int i) { S[i] = true; for (int j = 0; j < n; j++) { if (T[j]) continue; Type tmp = Lx[i] + Ly[j] - g[i][j]; if (!tmp) { T[j] = true; if (left[j] == -1 || dfs(left[j])) { left[j] = i; return true; } } else slack[j] = min(slack[j], tmp); } return false; } void update() { Type a = INF; for (int i = 0; i < n; i++) if (!T[i]) a = min(a, slack[i]); for (int i = 0; i < n; i++) { if (S[i]) Lx[i] -= a; if (T[i]) Ly[i] += a; } } int km() { for (int i = 0; i < n; i++) { left[i] = -1; Lx[i] = -INF; Ly[i] = 0; for (int j = 0; j < n; j++) Lx[i] = max(Lx[i], g[i][j]); } for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) slack[j] = INF; while (1) { for (int j = 0; j < n; j++) S[j] = T[j] = false; if (dfs(i)) break; else update(); } } int ans = 0; for (int i = 0; i < n; i++) ans -= g[left[i]][i]; return ans; } } gao; #define MP(a,b) make_pair(a,b) const int d[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}}; typedef pair<int, int> pii; const int N = 45; int t, n, m; char g[N][N]; int idb[N][N], idx[N][N], bn, xn, vis[N][N]; pii tob[N], tox[N]; int dis(pii a, pii b) { return abs(a.first - b.first) + abs(a.second - b.second); } void bfs(int xs, int ys) { queue<pii> Q; Q.push(MP(xs, ys)); memset(vis, INF, sizeof(vis)); vis[xs][ys] = 0; while (!Q.empty()) { pii u = Q.front(); if (g[u.first][u.second] == 'X') gao.add_Edge(idb[xs][ys], idx[u.first][u.second], -vis[u.first][u.second]); Q.pop(); for (int i = 0; i < 4; i++) { int x = u.first + d[i][0]; int y = u.second + d[i][1]; if (x < 0 || x >= n || y < 0 || y >= m || g[x][y] == '#') continue; if (vis[x][y] <= vis[u.first][u.second] + 1) continue; vis[x][y] = vis[u.first][u.second] + 1; Q.push(MP(x, y)); } } } int main() { scanf("%d", &t); while (t--) { scanf("%d%d", &n, &m); bn = xn = 0; memset(gao.g, -INF, sizeof(gao.g)); for (int i = 0; i < n; i++) { scanf("%s", g[i]); for (int j = 0; j < m; j++) { if (g[i][j] == 'B') { tob[bn] = MP(i, j); idb[i][j] = bn++; } if (g[i][j] == 'X') { tox[xn] = MP(i, j); idx[i][j] = xn++; } } } for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (g[i][j] == 'B') { bfs(i, j); } } } gao.init(bn); printf("%d\n", gao.km()); } return 0; }
UVA 10888 - Warehouse(二分图完美匹配)
标签:target std mod code tle [1] air ons memset
原文地址:http://www.cnblogs.com/yangykaifa/p/7097209.html