BZOJ2595 [Wc2008]游览计划

麻麻问我为什么跪倒在地

这么高端的求法！！！spfa优化DP。。。

等等，斯坦纳树的求法是DP？还是状压DP！Σ( ° △ °||)

蒟蒻彻底跪了，还是Orz hzwer吧2333

/**************************************************************
    Problem: 2595
    User: rausen
    Language: C++
    Result: Accepted
    Time:160 ms
    Memory:4444 kb
****************************************************************/
 
#include <cstdio>
#include <queue>
#include <utility>
 
#define P Point
using namespace std;
const int N = 15;
const int maxS = 1024;
const int inf = 1e9;
const int dx[4] = {0, 0, 1, -1};
const int dy[4] = {1, -1, 0, 0};
 
int n, m, K;
int a[N][N], bin[N];
int f[N][N][maxS];
bool inq[N][N], used[N][N];
 
struct Point {
    int x, y;
    P() {}
    P(int _x, int _y) : x(_x), y(_y) {}
};
queue<P> q;
 
struct Path {
    int x, y, s;
    Path() {}
    Path(int _f, int _s, int _t) : x(_f), y(_s), s(_t) {}
} path[N][N][maxS];
 
inline int read() {
    int x = 0;
    char ch = getchar();
    while (ch < '0' || '9' < ch)
        ch = getchar();
    while ('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x;
}
 
void spfa(int S) {
    int k, x, y, X, Y;
    while (!q.empty()) {
        inq[x = q.front().x][y = q.front().y] = 0, q.pop();
        for (k = 0; k < 4; ++k) {
            X = x + dx[k], Y = y + dy[k];
            if (x < 1 || y < 1 || x > n || y > m) continue;
            if (f[X][Y][S] > f[x][y][S] + a[X][Y]) {
                f[X][Y][S] = f[x][y][S] + a[X][Y];
                path[X][Y][S] = Path(x, y, S);
                if (!inq[X][Y])
                    q.push(P(X, Y)), inq[X][Y] = 1;
            }
        }
    }
}
 
#define t path[x][y][S]
void dfs(int x, int y, int S) {
    if (x > inf || !path[x][y][S].s) return;
    used[x][y] = 1;
    dfs(t.x, t.y, t.s);
    if (t.x == x && t.y == y) dfs(x, y, S ^ t.s);
}
#undef t
 
void read_in() {
    int i, j;
    n = read(), m = read();
    for (i = 1; i <= n; ++i)
        for (j = 1; j <= m; ++j)
            if (!(a[i][j] = read())) ++K;
    for (i = bin[0] = 1; i <= K; ++i)
        bin[i] = bin[i - 1] << 1;
}
 
void pre_work() {
    int i, j, k;
    for (i = 1; i <= n; ++i)
        for (j = 1; j <= m; ++j)
            for (k = 0; k < bin[K]; ++k)
                f[i][j][k] = path[i][j][k].x = inf;
    for (i = 1, K = 0; i <= n; ++i)
        for (j = 1; j <= m; ++j)
            if (!a[i][j])
                f[i][j][bin[K++]] = 0;
}
 
void work() {
    int i, j, S, s, tmp;
    for (S = 1; S < bin[K]; spfa(S++))
        for (i = 1; i <= n; ++i)
            for (j = 1; j <= m; ++j) {
                for (s = S & (S - 1); s; s = S & (s - 1))
                    if ((tmp = f[i][j][s] + f[i][j][S ^ s] - a[i][j]) < f[i][j][S]) {
                        f[i][j][S] = tmp;
                        path[i][j][S] = Path(i, j, s);
                    }
                if (f[i][j][S] != inf) {
                    q.push(P(i, j)), inq[i][j] = 1;
                }
            }
}
 
void get_ans1() {
    int i, j;
    for (i = 1; i <= n; ++i)
        for (j = 1; j <= m; ++j) if (!a[i][j]) {
            dfs(i, j, bin[K] - 1);
            printf("%d
", f[i][j][bin[K] - 1]);
            return;
        }
}
 
void get_ans2() {
    int i, j;
    for (i = 1; i <= n; ++i, putchar('
'))
        for (j = 1; j <= m; ++j)
            if (!a[i][j]) putchar('x'); else
            if (used[i][j]) putchar('o'); else
            putchar('_');
 
}
 
int main() {
    read_in();
    pre_work();
    work();
    get_ans1();
    get_ans2();
    return 0;
}

（p.s.  这英文是叫Steiner Tree吧。。。）