LuoGuP4294:[WC2008]游览计划

Pre

作为斯坦纳树的第一道题

Solution

以每一个格子建立点，然后直接跑斯坦纳树就可以了。

code

#include <cstdio>
#include <bitset>
#include <iostream>
#include <queue>
#include <limits.h>
#include <cstring>
#define ll long long
#define xx first
#define yy second
#define testname kkksc03
using namespace std;
const int N = 100 + 5, M = 10 + 5, mx = 2139062143, NN = 2800;
struct _in {
	const _in operator , (int &a) const {
		a = 0;
		char k = getchar ();
		int f = 1;
		while (k > '9' || k < '0') {
			if (k == '-') f = -1;
			k = getchar ();
		}
		while (k >= '0' && k <= '9') {
			a = a * 10 + k - '0';
			k = getchar ();
		}
		a *= f;
		return*this;
	}
};
inline int add (int u, int v) {
	if (u == mx || v == mx) return mx;
	else return u + v;
}
inline int mns (int u, int v) {
	return u == mx ? mx : u - v;
}
inline int min (int u, int v) {
	return u > v ? v : u;
}
inline int max (int u, int v) {
	return u > v ? u : v;
}
int dp[N][NN], n, m, mp[M][M];
inline int get (int x, int y) {
	return (x - 1) * m + y;
}
inline int GetLne (int x) {
	return (x - 1) / m + 1;
}
inline int GetCol (int x) {
	return (x - 1) % m + 1;
}
int anspos;
queue<int> q;
bool inq[N];
pair<int, int> s[N][NN];
inline void update (int u, int v, int w) {
	if (dp[v][w] > add (dp[u][w], mp[GetLne (v)][GetCol (v)])) {
		dp[v][w] = add (dp[u][w], mp[GetLne (v)][GetCol (v)]);
		s[v][w] = make_pair (u, w);
		if (!inq[v]) {
			inq[v] = 1;
			q.push (v);
		}
	}
}
inline void SPFA (int st) {
	while (!q.empty ()) {
		int tmp = q.front ();
		q.pop ();
		inq[tmp] = 0;
		int x = GetLne (tmp), y = GetCol (tmp);
		if (x != 1) update (tmp, tmp - m, st);
		if (x != n) update (tmp, tmp + m, st);
		if (y != 1) update (tmp, tmp - 1, st);
		if (y != m) update (tmp, tmp + 1, st);
	}
}
int k;
int op[M][M];
inline void dfs (int u, int v) {
//	printf ("%d ", u);
//	cout << bitset <8> (v) << " ";
//	printf ("%d ",s[u][v].xx);
//	cout << bitset <8> (s[u][v].yy) << endl;
	if (s[u][v].xx == -1) {
		int x = GetLne (u), y = GetCol (u);
		if (!mp[x][y]) op[x][y] = 'x';
		else op[x][y] = 'o';
		return ;
	}
	int x = GetLne (u), y = GetCol (u);
	op[x][y] = mp[x][y] ? 'o' : 'x';
	x = s[u][v].xx, y = s[u][v].yy;
	dfs (x, y);
	if (x == u) dfs (u, y ^ v);
}
int main () {
//	#ifdef chitongz
//	freopen ("x.in", "r", stdin);
//	freopen ("x.out", "w", stdout);
//	#endif
	memset (dp, 127, sizeof dp);
	memset (s, -1, sizeof s);
	scanf ("%d%d", &n, &m);
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			op[i][j] = '_';
			scanf ("%d", &mp[i][j]);
			if (!mp[i][j]) {
				anspos = get (i, j);
				++k;
				dp[get (i, j)][1 << (k - 1)] = 0;
			}
		}
	}
	int ans = INT_MAX;
	for (int j = 1; j <= (1 << k) - 1; ++j) {
		for (int i = 1; i <= n * m; ++i) {
			for (int tmp = j; tmp; tmp = (tmp - 1) & j) {
				if (dp[i][j] > mns (add (dp[i][tmp], dp[i][tmp ^ j]), mp[GetLne (i)][GetCol (i)])) {
					dp[i][j] = mns (add (dp[i][tmp], dp[i][tmp ^ j]), mp[GetLne (i)][GetCol (i)]);
					s[i][j] = make_pair (i, tmp);
				}
				if (dp[i][j] != dp[0][0]) {
					q.push (i);
					inq[i] = 1;
				}
			}
		}
		SPFA (j);
	}
	for (int i = 1; i <= n * m; ++i) {
		if (dp[i][(1 << k) - 1] < ans) {
			ans = dp[i][(1 << k) - 1];
			anspos = i;
		}
	}
	printf ("%d
", ans);
	dfs (anspos, (1 << k) - 1);
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			printf ("%c", op[i][j]);
		}
		puts ("");
	}
	return 0;
}

Conclusion

空间开小查错查了半天没有查出来，交上去发现 (M) 一般应该是 (10 + 5) ，我写的 (M=10)