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] = minlimits_{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] = minlimits_{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
", 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
*/