CF1155F Delivery Oligopoly

题意：给定简单无向图，求一个最小的边集使得这些点是边双，输出方案。n <= 14

解：考虑一个边双肯定是一条一条的链拼起来的。于是每次枚举一条链加上去就行了。

设f_s表示点集s形成边双的最小边数。link_i,j,s表示点集s能否形成一个i - j的链。link2_x,s表示点x和点集s是否直接相连。

上面这些数组都要记录方案，特别地，link2要记录两个方案，为了应对拼上去的链退化成一个点的情况。

#include <bits/stdc++.h>

const int N = 20000;

struct Edge {
    int nex, v;
}edge[N << 1]; int tp;

struct Node {
    int x, y, t;
    Node(int X = 0, int Y = 0, int T = 0) {
        x = X;
        y = Y;
        t = T;
    }
}fr3[N];

int pw[N], cnt[N], f[N], e[N], n, m;
bool link[20][20][N], link2[20][N];
int fr[20][20][N], fr2[20][N], fr22[20][N];

inline void add(int x, int y) {
    tp++;
    edge[tp].v = y;
    edge[tp].nex = e[x];
    e[x] = tp;
    return;
}

void out(int x, int y, int s) {
    if(cnt[s] == 1) return;
    printf("%d %d \n", y + 1, fr[x][y][s]);
    out(x, fr[x][y][s] - 1, s ^ (1 << y));
    return;
}

void out3(int s) {
    if(cnt[s] == 1) return;
    int x = fr3[s].x, y = fr3[s].y, t = fr3[s].t;
    out(x, y, t);
    printf("%d %d \n", x + 1, fr2[x][s ^ t]);
    if(x != y) printf("%d %d \n", y + 1, fr2[y][s ^ t]);
    else printf("%d %d \n", y + 1, fr22[y][s ^ t]);
    out3(s ^ t);
    return;
}

int main() {
    scanf("%d%d", &n, &m);
    for(int i = 1, x, y; i <= m; i++) {
        scanf("%d%d", &x, &y);
        add(x, y);
        add(y, x);
    }
    int lm = (1 << n) - 1; /// lm = 111111...1
    for(int i = 1; i <= lm; i++) {
        cnt[i] = 1 + cnt[i - (i & (-i))];
        if(i > 1) pw[i] = pw[i >> 1] + 1;
    }

    for(int x = 0; x < n; x++) {
        for(int s = 1; s <= lm; s++) {
            /// link2[x][s]
            if((s >> x) & 1) continue;
            for(int i = e[x + 1]; i; i = edge[i].nex) {
                int y = edge[i].v - 1;
                if((s >> y) & 1) {
                    link2[x][s] = 1;
                    if(!fr2[x][s]) {
                        fr2[x][s] = y + 1;
                    }
                    else {
                        fr22[x][s] = y + 1;
                        break;
                    }
                }
            }
        }
    }

    for(int i = 0; i < n; i++) {
        link[i][i][1 << i] = 1;
    }
    for(int s = 1; s < lm; s++) {
        for(int t1 = s, i; t1; t1 ^= (1 << i)) {
            i = pw[t1 & (-t1)];
            /// i + 1
            for(int t2 = s, x; t2; t2 ^= (1 << x)) {
                x = pw[t2 & (-t2)];
                /// f[i][x][s]
                if(!link[i][x][s]) continue;
                for(int j = e[x + 1]; j; j = edge[j].nex) {
                    int y = edge[j].v - 1;
                    if(((s >> y) & 1) == 0) {
                        link[i][y][s | (1 << y)] = 1;
                        fr[i][y][s | (1 << y)] = x + 1;
                    }
                }
            }
        }
    }

    memset(f, 0x3f, sizeof(f));
    f[1] = 0;
    for(int s = 2; s <= lm; s++) {
        /// f[s]
        for(int t = s & (s - 1); t; t = (t - 1) & s) {
            for(int t1 = t, x; t1; t1 ^= (1 << x)) {
                x = pw[t1 & (-t1)];
                for(int t2 = t, y; t2; t2 ^= (1 << y)) {
                    y = pw[t2 & (-t2)];
                    /// link[x][y][t] link2[x][s ^ t] link2[y][s ^ t]
                    if(link[x][y][t] && link2[x][s ^ t] && link2[y][s ^ t] && (x != y || fr22[x][s ^ t])) {
                        if(f[s] > f[s ^ t] + cnt[t] + 1) {
                            f[s] = f[s ^ t] + cnt[t] + 1;
                            fr3[s] = Node(x, y, t);
                        }
                    }
                }
            }
        }
    }

    printf("%d\n", f[lm]);
    out3(lm);
    return 0;
}