hdu 4640 Island and study-sister（最短路+状压dp）

题目链接：hdu 4640 Island and study-sister

解题思路

用二进制数表示2~n的点是否移动过的状态， dp[s][i] 表示状态s上的点必须经过并且当前在i节点的最小代价， 这步用类似最短路的方式求出。
然后是 dp2[i][s] 表示i个人移动过s状态的点的最小代价。

代码

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>

using namespace std;
const int maxn = 16;
const int maxs = (1<<maxn) + 5;
const int inf = 0x3f3f3f3f;
typedef pair<int,int> pii;

int N, M, W[maxs+5][maxn+2], val[maxs+5];
bool used[maxs+5][maxn+2];
vector<pii> G[maxn+5];

struct Node {
    int s, v, w;
    Node (int s = 0, int v = 0, int w = 0): s(s), v(v), w(w) {}
    bool operator < (const Node& u) const { return w > u.w; }
};

void presolve() {
    memset(W, inf, sizeof(W));
    memset(used, 0, sizeof(used));
    W[0][1] = 0;

    priority_queue<Node> que;
    que.push(Node(0, 1, W[0][1]));

    while (!que.empty()) {
        Node cur = que.top();
        que.pop();

        int s = cur.s;
        int u = cur.v;

        if (used[s][u]) continue;
        used[s][u] = true;

        for (int i = 0; i < G[u].size(); i++) {
            int v = G[u][i].first;
            int w = G[u][i].second;
            int vs = s;
            if (v > 1) vs |= (1<<(v-2));

            if (W[vs][v] > W[s][u] + w) {
                W[vs][v] = W[s][u] + w;
                que.push(Node(vs, v, W[vs][v]));
            }
        }
    }
}

void init () {
    scanf("%d%d", &N, &M);
    for (int i = 0; i <= N; i++) G[i].clear();

    int u, v, w;
    for (int i = 0; i < M; i++) {
        scanf("%d%d%d", &u, &v, &w);
        G[u].push_back(make_pair(v, w));
        G[v].push_back(make_pair(u, w));
    }

    presolve();
}

int dp[4][maxs+5];

int solve (int ed) {
    memset(dp, inf , sizeof(dp));
    dp[0][0] = 0;

    int as = (1<<(N-1))-1;
    for (int i = 0; i <= as; i++) {
        val[i] = inf;
        for (int j = 1; j <= N; j++)
            val[i] = min(val[i], W[i][j]);
    }

    for (int i = 0; i < 3; i++) {
        for (int s = 0; s <= as; s++) {
            if (dp[i][s] == inf) continue;
            int vs = as ^ s;
            for (int j = vs; j; j = (j-1)&vs)
                dp[i+1][j|s] = min(dp[i+1][j|s], max(dp[i][s], val[j]));
            dp[i+1][s] = min(dp[i+1][s], dp[i][s]);
        }
    }

    int ret = inf;
    for (int i = 0; i <= as; i++)
        if ((i&ed) == ed) ret =  min(ret, dp[3][i]);
    if (ret == inf) ret = -1;
    return ret;
}

int main () {
    int cas;
    scanf("%d", &cas);
    for (int kcas = 1; kcas <= cas; kcas++) {
        init();

        int s = 0, n, x;
        scanf("%d", &n);
        while (n--) {
            scanf("%d", &x);
            s |= (1<<(x-2));
        }

        printf("Case %d: %d\n", kcas, solve(s));
    }
    return 0;
}