POJ 2288 Islands and Bridges（状压DP）题解

题意：n个点，m有向边，w[i]表示i的价值，求价值最大的哈密顿图（只经过所有点一次）。价值为：所有点的w之和，加上，每条边的价值 = w[i] * w[j]，加上，如果连续的三个点相互连接的价值 = w[i] * w[j] * w[k]。n <= 13。

思路：dp[state][i][j]表示state状态下，最后两个为i，j。

代码：

#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include <iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double eps = 1e-8;
const int maxn = 15 + 10;
const int M = maxn * 30;
const ull seed = 131;
const int INF = 0x3f3f3f3f;
const int MOD = 1e4 + 7;
int w[maxn];
int n, m;
int g[maxn][maxn];
int dp[(1 << 13) + 10][maxn][maxn];
ll way[(1 << 13) + 10][maxn][maxn];
void solve(){
    memset(dp, -1, sizeof(dp));
    memset(way, 0, sizeof(way));
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++){
            if(i == j) continue;
            if(g[i][j] == INF) continue;
            dp[(1 << i) | (1 << j)][i][j] = w[i] + w[j] + g[i][j];
            way[(1 << i) | (1 << j)][i][j]++;
        }
    }
    for(int t = 0; t < (1 << n) - 1; t++){
        for(int i = 0; i < n; i++){
            if(!((1 << i) & t)) continue;
            for(int j = 0; j < n; j++){
                if(!((1 << j) & t)) continue;
                if(g[i][j] == INF) continue;
                if(dp[t][i][j] == -1) continue;
                for(int k = 0; k < n; k++){
                    if((1 << k) & t) continue;
                    if(g[j][k] == INF) continue;
                    ll ret = dp[t][i][j] + w[k] + g[j][k];
                    if(g[i][k] != INF) ret += w[i] * w[j] * w[k];
                    if(dp[(1 << k) | t][j][k] < ret){
                        dp[(1 << k) | t][j][k] = ret;
                        way[(1 << k) | t][j][k] = way[t][i][j];
                    }
                    else if(dp[(1 << k) | t][j][k] == ret){
                        way[(1 << k) | t][j][k] += way[t][i][j];
                    }
                }
            }
        }
    }
}
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        scanf("%d%d", &n, &m);
        for(int i = 0; i < n; i++)
            scanf("%d", &w[i]);
        memset(g, INF, 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] = w[u] * w[v];
        }
        if(n == 1){
            printf("%d 1\n", w[0]);
            continue;
        }
        solve();
        ll ans = 0, num = 0;
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                if(i == j) continue;
                if(g[i][j] == INF) continue;
                if(dp[(1 << n) - 1][i][j] > ans){
                    ans = dp[(1 << n) - 1][i][j];
                    num = way[(1 << n) - 1][i][j];
                }
                else if(dp[(1 << n) - 1][i][j] == ans){
                    num += way[(1 << n) - 1][i][j];
                }
            }
        }
        printf("%lld %lld\n", ans, num / 2);
    }
    return 0;
}
/*
3
3 1
2 2 2
1 2
*/

原文地址：https://www.cnblogs.com/KirinSB/p/10945242.html