【HDOJ】3311 Dig The Wells

Steiner Tree。概念就不讲了，引入0号结点。[1, n+m]到0连一条边，权重表示挖井的费用。这样建图spfa求MST即满足所求解。

 /* 3311 */
 #include <iostream>
 #include <string>
 #include <map>
 #include <queue>
 #include <set>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <algorithm>
 #include <cstdio>
 #include <cmath>
 #include <ctime>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cassert>
 #include <functional>
 #include <iterator>
 #include <iomanip>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,1024000")

 #define sti                set<int>
 #define stpii            set<pair<int, int> >
 #define mpii            map<int,int>
 #define vi                vector<int>
 #define pii                pair<int,int>
 #define vpii            vector<pair<int,int> >
 #define rep(i, a, n)     for (int i=a;i<n;++i)
 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 #define clr                clear
 #define pb                 push_back
 #define mp                 make_pair
 #define fir                first
 #define sec                second
 #define all(x)             (x).begin(),(x).end()
 #define SZ(x)             ((int)(x).size())
 #define lson            l, mid, rt<<1
 #define rson            mid+1, r, rt<<1|1

 typedef struct {
     int v, w, nxt;
 } edge_t;

 const int INF = 0x3f3f3f3f;
 const int maxst = ;
 const int maxn = ;
 const int maxe = ;
 edge_t E[maxe];
 int head[maxn], l;
 int dp[maxn][maxst];
 bool visit[maxn][maxst];
 int n, m;
 queue<pii> Q;

 void init() {
     l = ;
     memset(head, -, sizeof(head));
     memset(dp, INF, sizeof(dp));
     memset(visit, false, sizeof(visit));
 }

 void addEdge(int u, int v, int w) {
     E[l].v = v;
     E[l].w = w;
     E[l].nxt = head[u];
     head[u] = l++;

     E[l].v = u;
     E[l].w = w;
     E[l].nxt = head[v];
     head[v] = l++;
 }

 void spfa() {
     int u, v, st, nst;

     while (!Q.empty()) {
         u = Q.front().fir;
         st = Q.front().sec;
         visit[u][st] = false;
         Q.pop();
         for (int i=head[u]; i!=-; i=E[i].nxt) {
             v = E[i].v;
             nst = st;
             if (v <= n)
                 nst |= (<<v);
             if (dp[u][st]+E[i].w < dp[v][nst]) {
                 dp[v][nst] = dp[u][st]+E[i].w;
                 if (nst==st && !visit[v][nst]) {
                     visit[v][nst] = true;
                     Q.push(mp(v, nst));
                 }
             }
         }
     }
 }

 void solve() {
     int mst = <<(n+);
     int nn = n + m + ;
     int kk, kk_;

     rep(i, , n+)
         dp[i][<<i] = ;

     rep(i, , mst) {
         rep(j, , nn) {
             rep(k, , i) {
                 if ((k|i) == i) {
                     kk = k;
                     kk_ = i - k;
                     if (j <= n) {
                         kk |= (<<j);
                         kk_ |= (<<j);
                     }
                     dp[j][i] = min(dp[j][i], dp[j][kk]+dp[j][kk_]);
                 }
             }
             if (dp[j][i] != INF) {
                 Q.push(mp(j, i));
                 visit[j][i] = true;
             }
         }
         spfa();
     }

     int ans = INF;
     rep(j, , nn)
         ans = min(ans, dp[j][mst-]);
     printf("%d\n", ans);
 }

 int main() {
     ios::sync_with_stdio(false);
     #ifndef ONLINE_JUDGE
         freopen("data.in", "r", stdin);
         freopen("data.out", "w", stdout);
     #endif

     int p;
     int u, v, w;

     while (scanf("%d %d %d",&n,&m,&p)!=EOF) {
         init();
         rep(i, , n+m+) {
             scanf("%d", &w);
             addEdge(, i, w);
         }
         rep(i, , p) {
             scanf("%d %d %d", &u, &v, &w);
             addEdge(u, v, w);
         }
         solve();
     }

     #ifndef ONLINE_JUDGE
         printf("time = %d.\n", (int)clock());
     #endif

     return ;
 }