#333. 【NOIP2017】宝藏
1、错误的$n^42^n$做法：
dp[s]表示当前的点集为s，然后从这些点中选一个做起点i，然后枚举边，然后更新dp[t|(1<<j)]。dis[s][i]表示点集为s的情况下的i号点的深度。详见代码。
为什么是错的，不满足当前最优一定是最后最优。比如下面的hack数据，正确答案应该是10420，即从4号点出发，长度为1的那条边不要。而在上面的dp中，点集为234的状态中只能从24和34转移而来，而这两个取最小值的时候，一定会算上1这条边，答案为12，而不是10这条边。导致下一步dp的时候，深度边长，导致更不优。
1 #include<cstdio> 2 #include<algorithm> 3 #include<iostream> 4 #include<cstring> 5 #include<cctype> 6 #include<cmath> 7 #include<set> 8 #include<queue> 9 #include<vector> 10 #include<map> 11 #include<bitset> 12 using namespace std; 13 typedef long long LL; 14 15 //char buf[100000], *p1 = buf, *p2 = buf; 16 //#define nc() p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++ 17 18 inline int read() { 19 int x = 0, f = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch=='-') f = -1; 20 for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0'; return x * f; 21 } 22 23 const int N = 100; 24 const LL LLINF = 1e18; 25 const int INF = 1e9; 26 27 LL dp[(1 << 12) + 10]; 28 int mp[N][N], dis[(1 << 12) + 10][15]; 29 int n, m; 30 31 LL solve(int x) { 32 memset(dis, 0, sizeof(dis)); 33 int S = (1 << n) - 1; 34 35 for (int s = 0; s <= S; ++s) dp[s] = LLINF; 36 dis[1 << x][x] = 1, dp[1 << x] = 0; 37 38 for (int s = 0, t; s <= S; ++s) { // 点集 39 if (dp[s] == LLINF) continue; 40 cout << bitset<6>(s) << " "; 41 for (int i = 0; i < n; ++i) { // 选一个点当起点 42 if (!((s >> i) & 1)) continue; 43 for (int j = 0; j < n; ++j) { // 到达的下一个点 44 if (i == j || mp[i][j] == INF || ((s >> j) & 1)) continue; 45 t = s | (1 << j); 46 cout << bitset<6>(t) << " "; 47 if (dp[t] > dp[s] + dis[s][i] * mp[i][j]) { 48 dp[t] = dp[s] + dis[s][i] * mp[i][j]; 49 for (int k = 0; k < n; ++k) dis[t][k] = dis[s][k]; 50 dis[t][j] = dis[s][i] + 1; 51 cout << dp[t] << " "; 52 } 53 } 54 } 55 } 56 return dp[S]; 57 } 58 59 int main() { 60 // freopen("1.txt", "r", stdin); 61 // freopen("2.txt", "w", stdout); 62 n = read(), m = read(); 63 64 for (int i = 0; i <= n; ++i) 65 for (int j = 0; j <= n; ++j) mp[i][j] = INF; 66 for (int i = 1; i <= m; ++i) { 67 int u = read() - 1, v = read() - 1, w = read(); 68 mp[u][v] = min(mp[u][v], w), mp[v][u] = min(mp[v][u], w); 69 } 70 71 LL ans = 1e18; 72 for (int i = 0; i < n; ++i) 73 ans = min(ans, solve(i)); 74 cout << ans; 75 76 return 0; 77 } 78 /* 79 80 hack数据： 81 dp不满足，当前最优，不满足最后的答案最优 82 83 6 14 84 1 2 101 85 1 3 44 86 1 4 235 87 1 6 629 88 2 3 60 89 2 4 196 90 2 5 250 91 2 6 490 92 3 4 237 93 3 5 114 94 3 6 715 95 4 5 166 96 4 6 432 97 5 6 932 98 99 6 6 100 1 2 100 101 2 3 1 102 2 4 10 103 3 4 10 104 3 5 100 105 4 6 10000 106 107 */
2、那么解决上述dp中出现的dp，就需要按深度进行dp，每次枚举下一深度的所有点，子集dp。复杂度$n^33^n$
1 #include<cstdio> 2 #include<algorithm> 3 #include<iostream> 4 #include<cstring> 5 #include<cctype> 6 #include<cmath> 7 #include<set> 8 #include<queue> 9 #include<vector> 10 #include<map> 11 #include<bitset> 12 using namespace std; 13 typedef long long LL; 14 15 inline int read() { 16 int x = 0, f = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch=='-') f = -1; 17 for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0'; return x * f; 18 } 19 20 const int N = 100; 21 const LL LLINF = 1e18; 22 const int INF = 1e9; 23 24 LL dp[(1 << 12) + 10][13], tmp[(1 << 12) + 10], dis[N]; 25 int mp[N][N]; 26 int n, m; 27 28 LL solve(int x) { 29 int S = (1 << n) - 1; 30 for (int s = 0; s <= S; ++s) 31 for (int i = 0; i <= n; ++i) dp[s][i] = LLINF; 32 dp[1 << x][1] = 0; 33 34 for (int s = 0; s <= S; ++s) { 35 int r = 0; 36 for (int i = 0; i < n; ++i) dis[i] = LLINF; 37 for (int i = 0; i < n; ++i) { 38 if ((s >> i) & 1) 39 for (int j = 0; j < n; ++j) 40 if (!((s >> j) & 1) && mp[i][j]) 41 r |= (1 << j), dis[j] = min(dis[j], (LL)mp[i][j]); 42 } 43 for (int t = r; t; t = (t - 1) & r) { 44 tmp[t] = 0; 45 for (int i = 0; i < n; ++i) 46 if ((t >> i) & 1) tmp[t] += dis[i]; 47 } 48 for (int d = 1; d <= n; ++d) { 49 if (dp[s][d] == LLINF) continue; 50 for (int t = r; t; t = (t - 1) & r) 51 dp[s | t][d + 1] = min(dp[s | t][d + 1], dp[s][d] + tmp[t] * d); 52 } 53 } 54 LL ans = LLINF; 55 for (int i = 0; i <= n; ++i) ans = min(ans, dp[S][i]); 56 return ans; 57 } 58 59 int main() { 60 61 n = read(), m = read(); 62 63 for (int i = 0; i <= n; ++i) 64 for (int j = 0; j <= n; ++j) mp[i][j] = INF; 65 for (int i = 1; i <= m; ++i) { 66 int u = read() - 1, v = read() - 1, w = read(); 67 mp[u][v] = min(mp[u][v], w), mp[v][u] = min(mp[v][u], w); 68 } 69 70 LL ans = 1e18; 71 for (int i = 0; i < n; ++i) 72 ans = min(ans, solve(i)); 73 cout << ans; 74 75 return 0; 76 }
3、随机化算法