2019ICPC南昌邀请赛现场赛A题 - Attack(斯坦纳树)

题意：

给出一张图，求让\(4\)对点相互可以到达的最小边权值。仅要求一对之间，一对与另外一对可到达也可不到达。

分析：

斯坦纳树裸题，众所周知斯坦纳树仅能求出这\(4\)对点(关键点)的连通状况，如这\(4\)对点相互都连通，某点和某点连通等。然而让这\(4\)对点连通符合题目要求，但不一定是最优解(我可以让每对点直接相连)，所以我们要对斯坦纳树求出的\(dp\)数组进行子集\(dp\)才能得到最优解。

#include <bits/stdc++.h>
using namespace std;

const int N  = 50 + 5, M = 1000 + 5;
const int inf = 0x3f3f3f3f;

int n, m, w, tot, cnt, maxsta, head[N], vis[N];
int dp[N][1 << 10], ans[1 << 11];
string a, b;
queue<int> q;
map<string, int> maps;

struct node {
    int v, w, next;
} e[M << 1];

void init() {
    tot = cnt = 0;
    memset(dp, inf, sizeof dp);
    memset(head, 0, sizeof head);
    memset(vis, 0, sizeof vis);
    while (!q.empty()) q.pop();
    maps.clear();
}

void addedge(int u, int v, int w) {
    e[++cnt].v = v;
    e[cnt].w = w;
    e[cnt].next = head[u];
    head[u] = cnt;
}

void spfa(int sta) {
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        vis[u] = 0;
        for (int i = head[u]; i; i = e[i].next) {
            int v = e[i].v, val = e[i].w;
            if (dp[v][sta] > dp[u][sta] + val) {
                dp[v][sta] = dp[u][sta] + val;
                if (!vis[v]) q.push(v), vis[v] = 1;
            }
        }
    }
}

bool check(int sta) {
    for (int i = 0; i < 8; i += 2) {
        if ((sta >> i & 1) ^ (sta >> (i + 1) & 1)) return false;
    }
    return true;
}

int main() {
    while (cin >> n >> m) {
        init();
        for (int i = 1; i <= n; i++) {
            cin >> a;
            maps[a] = ++tot;
        }
        for (int i = 1; i <= m; i++) {
            cin >> a >> b >> w;
            addedge(maps[a], maps[b], w);
            addedge(maps[b], maps[a], w);
        }
        for (int i = 0; i < 8; i++) {
            cin >> a;
            dp[maps[a]][1 << i] = 0;
        }
        maxsta = 1 << 8;
        for (int sta = 0; sta < maxsta; sta++) {
            while(!q.empty()) q.pop();
            for (int i = 1; i <= n; i++) {
                for (int s = sta; s; s = (s - 1) & sta) {
                    if(dp[i][sta] > dp[i][s] + dp[i][sta ^ s])
                        dp[i][sta] = dp[i][s] + dp[i][sta ^ s];
                }
                if(dp[i][sta] < inf) q.push(i), vis[i] = 1;
            }
            spfa(sta);
        }
        memset(ans, inf, sizeof ans);
        for (int s = 0; s < maxsta; s++) {
            if (!check(s)) continue;
            for (int i = 1; i <= n; i++) {
                ans[s] = min(ans[s], dp[i][s]);
            }
        }
        for (int s = 0; s < maxsta; s++) {
            if (!check(s)) continue;
            for (int p = (s - 1) & s; p; p = (p - 1) & s)
                ans[s] = min(ans[s], ans[p] + ans[s ^ p]);
        }
        cout << ans[maxsta - 1] << '\n';
    }
    return 0;
}

原文地址：https://www.cnblogs.com/ChaseNo1/p/11644939.html