POJ 2288 Islands and Bridges（状压DP）

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题目链接：http://poj.org/problem?id=2288

题意：仍然是求一条哈密顿回路，不过权值和的计算包括三部分：所经过的点权和，连续经过的两个点的权值乘积，连续三个点彼此可达则再加上三点乘积，问最大权值为多少，且有多少种方案达到该权值，注意一条路径的走法只能算一次

思路：dp[sta][i][j]表示当前状态为sta，前一个访问点为i，当前访问点为j所获得的权值，num[sta][i][j]相应对应方案数，转移即可

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <utility>#include <cmath>#include <queue>#include <set>#include <map>#include <climits>#include <functional>#include <deque>#include <ctime>#include <string>#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;typedef long long ll;int n, m;int g[13][13], val[13];ll dp[1 << 13][13][13], num[1 << 13][13][13];void init(){memset(dp, -1, sizeof(dp));memset(num, 0, sizeof(num));for (int i = 0; i < n; i++)for (int j = 0; j < n; j++){if (i == j || !g[i][j]) continue;dp[(1 << i) | (1 << j)][i][j] = val[i] + val[j] + val[i] * val[j];num[(1 << i) | (1 << j)][i][j] = 1;}}void solve(){for (int i = 0; i < (1 << n); i++){for (int j = 0; j < n; j++){if (i & (1 << j)){for (int k = 0; k < n; k++){if (j == k || !g[j][k] || (i & (1 << k)) == 0 || dp[i][j][k] == -1) continue;for (int l = 0; l < n; l++){if (j == l || k == l || !g[k][l] || (i & (1 << l))) continue;ll w = dp[i][j][k] + val[l] + val[k] * val[l];if (g[j][l]) w += val[j] * val[k] * val[l];if (dp[i | (1 << l)][k][l] < w){dp[i | (1 << l)][k][l] = w;num[i | (1 << l)][k][l] = num[i][j][k];}else if (dp[i | (1 << l)][k][l] == w)num[i | (1 << l)][k][l] += num[i][j][k];}}}}}}int main(){int t;cin >> t;while (t--){scanf("%d%d", &n, &m);for (int i = 0; i < n; i++)scanf("%d", &val[i]);memset(g, 0, sizeof(g));for (int i = 0; i < m; i++){int u, v;scanf("%d%d", &u, &v);u--, v--;g[u][v] = g[v][u] = 1;}if (n == 1){cout << val[0] << " " << 1 << endl;continue;}init();solve();ll ans1 = 0, ans2 = 0;for (int i = 0; i < n; i++)for (int j = 0; j < n; j++){if (i == j || !g[i][j]) continue;if (ans1 < dp[(1 << n) - 1][i][j]){ans1 = dp[(1 << n) - 1][i][j];ans2 = num[(1 << n) - 1][i][j];}else if (ans1 == dp[(1 << n) - 1][i][j])ans2 += num[(1 << n) - 1][i][j];}cout << ans1 << " " << ans2 / 2 << endl;}return 0;}

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