【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 = 65;
const int maxn = 1010;
const int maxe = 13005;
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 = 0;
    memset(head, -1, 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!=-1; i=E[i].nxt) {
            v = E[i].v;
            nst = st;
            if (v <= n)
                nst |= (1<<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 = 1<<(n+1);
    int nn = n + m + 1;
    int kk, kk_;
    
    rep(i, 0, n+1)
        dp[i][1<<i] = 0;
    
    rep(i, 0, mst) {
        rep(j, 0, nn) {
            rep(k, 1, i) {
                if ((k|i) == i) {
                    kk = k;
                    kk_ = i - k;
                    if (j <= n) {
                        kk |= (1<<j);
                        kk_ |= (1<<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, 0, nn)
        ans = min(ans, dp[j][mst-1]);
    printf("%d
", 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, 1, n+m+1) {
            scanf("%d", &w);
            addEdge(0, i, w);
        }
        rep(i, 0, p) {
            scanf("%d %d %d", &u, &v, &w);
            addEdge(u, v, w);
        }
        solve();
    }
    
    #ifndef ONLINE_JUDGE
        printf("time = %d.
", (int)clock());
    #endif
    
    return 0;
}