poj 2288 Islands and Bridges (用DP解汉密尔顿问题)
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Islands and Bridges
Description
Given a map of islands and bridges that connect these islands, a Hamilton path, as we all know, is a path along the bridges such that it visits each island exactly once. On our map, there is also a positive integer value associated with each island. We call a Hamilton path the best triangular Hamilton path if it maximizes the value described below.
Suppose there are n islands. The value of a Hamilton path C1C2...Cn is calculated as the sum of three parts. Let Vi be the value for the island Ci. As the first part, we sum over all the Vi values for each island in the path. For the second part, for each edge CiC i+1 in the path, we add the product Vi*V i+1. And for the third part, whenever three consecutive islands CiC i+1C i+2 in the path forms a triangle in the map, i.e. there is a bridge between Ci and C i+2, we add the product Vi*V i+1*V i+2. Most likely but not necessarily, the best triangular Hamilton path you are going to find contains many triangles. It is quite possible that there might be more than one best triangular Hamilton paths; your second task is to find the number of such paths. Input
The input file starts with a number q (q<=20) on the first line, which is the number of test cases. Each test case starts with a line with two integers n and m, which are the number of islands and the number of bridges in the map, respectively. The next line contains n positive integers, the i-th number being the Vi value of island i. Each value is no more than 100. The following m lines are in the form x y, which indicates there is a (two way) bridge between island x and island y. Islands are numbered from 1 to n. You may assume there will be no more than 13 islands.
Output
For each test case, output a line with two numbers, separated by a space. The first number is the maximum value of a best triangular Hamilton path; the second number should be the number of different best triangular Hamilton paths. If the test case does not contain a Hamilton path, the output must be `0 0'.
Note: A path may be written down in the reversed order. We still think it is the same path. Sample Input 2 3 3 2 2 2 1 2 2 3 3 1 4 6 1 2 3 4 1 2 1 3 1 4 2 3 2 4 3 4 Sample Output 22 3 69 1 Source
Shanghai 2004
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题意：
求一条哈密顿回路，但是权值计算不同，包括三部分：
1.经过个所有点的权值相加。
2.经过的连续两个点的权值的乘积。
3，能够构成三角型的连续三个点的乘积。
这些全部加起来就是这条回路的总权值。输出最大权值和这个最大权值的路线有多少条。
思路：
三维DP：
第一维：二进制状态压缩存访问的岛屿的情况。
第二维：存上一个点访问的岛屿标号。
第三维：存上上个点访问的岛屿标号。
注意：
要考虑只有一个岛屿的情况。还有一条汉密尔顿路径在无向图中实际上为两条，但题目只算作一条。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#define maxn 13
#define maxx 1<<13
using namespace std;
typedef long long ll;
int n,m;
bool vis[maxn][maxn];
ll ans,cnt;
ll dp[maxx][maxn][maxn]; // 状态、上一次访问的岛屿、上上次
ll num[maxx][maxn][maxn];
ll val[maxn];
vector<int>edge[maxn];
void init()
{
int i,j,u,v,s,s1,s2;
memset(dp,-1,sizeof(dp));
memset(num,0,sizeof(num));
memset(vis,0,sizeof(vis));
for(i=0; i<n; i++)
{
edge[i].clear();
}
for(i=1; i<=m; i++)
{
scanf("%d%d",&u,&v);
u--,v--;
vis[u][v]=vis[v][u]=1;
edge[u].push_back(v);
edge[v].push_back(u);
s1=1<<u;
s2=1<<v;
s=0|s1|s2;
dp[s][u][v]=dp[s][v][u]=val[u]+val[v]+val[u]*val[v];
num[s][u][v]=num[s][v][u]=1;
}
}
void solve()
{
int i,j,k,p,r,s,sz,ss,v;
ll nv,nnum,tv;
r=1<<n;
for(k=0; k<r; k++) // 对状态循环
{
s=k;
for(i=0; i<n; i++) // 对上一次循环
{
if(s&(1<<i)) // 如果s能把i当做上次
{
for(j=0; j<n; j++) // 对上上循环
{
if(i==j||!(s&(1<<j))||dp[s][i][j]==-1) continue ;
sz=edge[i].size();
for(p=0; p<sz; p++)
{
v=edge[i][p];
if(s&(1<<v)) continue ;
nv=dp[s][i][j];
nnum=num[s][i][j];
tv=nv+val[v]+val[i]*val[v];
if(vis[j][v]) tv+=val[j]*val[i]*val[v];
ss=s|(1<<v);
if(dp[ss][v][i]<tv) // 没出现则更新
{
dp[ss][v][i]=tv;
num[ss][v][i]=nnum;
}
else if(dp[ss][v][i]==tv) // 出现累加num
{
num[ss][v][i]+=nnum;
}
}
}
}
}
}
}
int main()
{
int i,j,t,s;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
for(i=0; i<n; i++)
{
scanf("%I64d",&val[i]);
}
init();
if(n==1)
{
printf("%I64d 1\n",val[0]);
continue ;
}
solve();
ans=-1;
s=(1<<n)-1;
for(i=0; i<n; i++)
{
for(j=0; j<n; j++)
{
if(ans<dp[s][i][j]) ans=dp[s][i][j];
}
}
if(ans==-1)
{
printf("0 0\n");
continue ;
}
cnt=0;
for(i=0; i<n; i++)
{
for(j=0; j<n; j++)
{
if(ans==dp[s][i][j]) cnt+=num[s][i][j];
}
}
printf("%I64d %I64d\n",ans,cnt/2);
}
return 0;
}