[题目链接]
[算法]
状压DP
f[i][j][S]表示j的深度为i,要从第j个宝藏屋开始挖,挖出集合S的最小代价
有状态转移方程 :
f[i][j][S] = min{ (i + 1) * dist(j,k) + f[i + 1][k][S2 - {k}] + f[i][j][S - S2] } ( 其中,S2为S的子集)
具体实现时需要枚举子集
[代码]
#include<bits/stdc++.h> using namespace std; #define MAXN 12 const int MAXS = 1 << 12; const int INF = 1e9; int i,j,k,mask,cost,u,v,w,t1,t2,ans,n,m,S,S2; int dist[MAXN][MAXN]; int f[MAXN][MAXN][MAXS]; int bit[MAXN],size[MAXS]; int main() { scanf("%d%d",&n,&m); for (i = 0; i < n; i++) { for (j = 0; j < n; j++) { dist[i][j] = INF; } } for (i = 1; i <= m; i++) { scanf("%d%d%d",&u,&v,&w); u--; v--; dist[u][v] = dist[v][u] = min(dist[u][v],w); } mask = (1 << n) - 1; for (i = 0; i < n; i++) { for (j = 0; j < n; j++) { for (k = 0; k <= mask; k++) { f[i][j][k] = INF; } } } bit[0] = 1; for (i = 1; i < n; i++) bit[i] = bit[i - 1] << 1; for (i = 1; i < n; i++) size[i] = size[(i - 1) & i] + 1; for (i = 0; i < n; i++) f[n - 1][i][0] = 0; for (i = 0; i <= n - 2; i++) { for (j = 0; j < n; j++) { f[i][j][0] = 0; } } for (i = n - 2; i >= 0; i--) { for (j = 0; j < n; j++) { for (S = 1; S <= mask; S++) { if (size[S] > n - i - 1) continue; for (k = 0; k < n; k++) { if (dist[j][k] == INF) continue; cost = dist[j][k] * (i + 1); v = INF; if (!(S & bit[k])) continue; for (S2 = S; S2; S2 = (S2 - 1) & S) { if (!(S2 & (1 << k))) continue; t1 = f[i + 1][k][S2 ^ bit[k]]; t2 = f[i][j][S ^ S2]; if (t1 + t2 < v) v = t1 + t2; } if (cost + v < f[i][j][S]) f[i][j][S] = cost + v; } } } } ans = INF; for (i = 0; i < n; i++) ans = min(ans,f[0][i][mask ^ bit[i]]); printf("%d ",ans); return 0; }