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

来源：互联网发布：php源码分析弱类型编辑：程序博客网时间：2024/09/21 06:21

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;}

0 0