[WC2008]游览计划 「斯坦那树模板」

斯坦那树

百度释义

斯坦纳树问题是组合优化问题，与最小生成树相似，是最短网络的一种。最小生成树是在给定的点集和边中寻求最短网络使所有点连通。而最小斯坦纳树允许在给定点外增加额外的点，使生成的最短网络开销最小。

即最小斯坦那树即为并非选择所有的结点，而是选择一部分结点，为保证它们连通，且求解最小开销

题解

斯坦那树模板

发现直接表示点的存在性没有意义

设函数 \(f[i][state]\) 表示：对于点 \(i\)，其它结点与其连通情况

那么有两种转移

其一、由其子集转移

\[f[i][state] = \min\limits_{sub \in state} \{f[i][sub] + f[i][\complement_{state}sub] - value_i\}\]

之所以要减去 \(value_i\) 是因为会算重

附：枚举子集的方法

for (int sub = state & (state - 1); sub; sub = (sub - 1) & state)

其二、由相邻当前状态下结点转移

\[f[i][state] = \min\limits_{state_p = true} \{f[p][state] + value_i\}\]

发现很像三角形不等式，故考虑 \(SPFA\) 转移

总复杂度 \(O (n3^n + kE2^n)\)，\(3^n\) 为枚举子集总复杂度

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;

const int MAXN = 10 + 5;
const int MAXM = 1 << 10;

const int INF = 0x3f3f3f3f;

const int NextX[4]= {- 1, 0, 0, 1}, NextY[4]= {0, - 1, 1, 0};

int N, M;
int Map[MAXN][MAXN]= {0};

struct preSt {
    int x, y;
    int state;

    preSt (int fx = 0, int fy = 0, int fs = 0) :
        x (fx), y (fy), state (fs) {}
} ;

int f[MAXN][MAXN][MAXM];
preSt pre[MAXN][MAXN][MAXM];
int cnt = 0;

queue<pair<int, int> > que;
void SPFA (int state) {
    while (! que.empty()) {
        pair<int, int> top = que.front();
        que.pop();

        int x = top.first, y = top.second;
        for (int i = 0; i < 4; i ++) {
            int tx = x + NextX[i];
            int ty = y + NextY[i];
            if (tx < 1 || tx > N || ty < 1 || ty > M)
                continue;
            if (f[x][y][state] + Map[tx][ty] < f[tx][ty][state]) {
                f[tx][ty][state] = f[x][y][state] + Map[tx][ty];
                pre[tx][ty][state] = preSt (x, y, state);
                que.push(make_pair (tx, ty));
            }
        }
    }
}

int tag[MAXN][MAXN]= {0};
void traceback (int x, int y, int state) {
    if (! x || ! y)
        return ;
    tag[x][y] = 1;
    preSt pr = pre[x][y][state];
    traceback (pr.x, pr.y, pr.state);
    if (pr.x == x && pr.y == y)
        traceback (pr.x, pr.y, state - pr.state);
}

int getnum () {
    int num = 0;
    char ch = getchar ();

    while (! isdigit (ch))
        ch = getchar ();
    while (isdigit (ch))
        num = (num << 3) + (num << 1) + ch - '0', ch = getchar ();

    return num;
}

int main () {
    memset (f, 0x3f, sizeof (f));
    N = getnum (), M = getnum ();
    int px, py;
    for (int i = 1; i <= N; i ++)
        for (int j = 1; j <= M; j ++) {
            Map[i][j] = getnum ();
            if (! Map[i][j]) {
                cnt ++, f[i][j][1 << (cnt - 1)] = 0;
                px = i, py = j;
            }
        }
    int limit = (1 << cnt) - 1;
    for (int state = 1; state <= limit; state ++) {
        for (int i = 1; i <= N; i ++)
            for (int j = 1; j <= M; j ++) {
                for (int sub = state & (state - 1); sub; sub = (sub - 1) & state) // from subset
                    if (f[i][j][sub] + f[i][j][state - sub] - Map[i][j] < f[i][j][state]) {
                        f[i][j][state] = f[i][j][sub] + f[i][j][state - sub] - Map[i][j];
                        pre[i][j][state] = preSt (i, j, sub);
                    }
                if (f[i][j][state] < INF)
                    que.push(make_pair (i, j));
            }
        SPFA (state); // from other nodes
    }
    traceback (px, py, limit);
    printf ("%d\n", f[px][py][limit]);
    for (int i = 1; i <= N; i ++) {
        for (int j = 1; j <= M; j ++) {
            if (! Map[i][j])
                putchar ('x');
            else {
                tag[i][j] ? putchar ('o') : putchar ('_');
            }
        }
        puts ("");
    }

    return 0;
}

/*
4 4
0 1 1 0
2 5 5 1
1 5 5 1
0 1 1 0
*/

原文地址：https://www.cnblogs.com/Colythme/p/10328442.html