Islands and Bridges
| Time Limit: 4000MS | Memory Limit: 65536K | |
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
CiCi+1 in the path, we add the product Vi*Vi+1. And for the third part, whenever three consecutive islands CiCi+1Ci+2 in the path forms a triangle in the map, i.e. there is a bridge between Ci and Ci+2,
we add the product Vi*Vi+1*Vi+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
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题意：给出n个点，m条边。每个点有一个权值w。找出一条汉密尔顿路径，使它的值最大。一条汉密尔顿路径的值由三部分组成：
1) 路径上每个点的权值之和
2) 路径上每条边u-v，将其权值的积累加起来。即w[u]*w[v]
3) 如果三个点形成一个三角形，例如i、i+1、i+2，那么将w[i]*w[i+1]*w[i+2]累加起来
一条汉密尔顿路径可能包含多个三角形，一张图中也可能包含多个最好的汉密尔顿路径。输出最大的汉密尔顿路径的值，以及这样的汉密尔顿路径的个数。同一条汉密尔顿路径的两种走法算作一种。
思路：汉密尔顿：经过每个点一次且仅一次。
汉密尔顿回路问题是旅行商问题。但是别想多了，这种NP完全问题大数据无解。而且这不是回路，是路径。
注意数据范围，13。这就是摆明了让你进行状态压缩DP的。
状态像这样描述：
dp[p][i][j](dp[1<<13][13][13]): p代表当前已经拜访的点的状态，j是前一个点，i是前一个点的前一个点。
ways[p][i][j]: 储存当前最好的汉密尔顿路径的条数。
我们的目标是dp[1<<13-1][x][y]，因此也不需要判断是否存在汉密尔顿路径（况且至今没找到等价条件）
首先是初始化（在判断三角形之前 我们可以这样初始化）：
dp[1<<i|1<<j][i][j] = w[i] + w[j] + w[i] * w[j]
ways[1<<i|1<<j][i][j] = 1
状态转移：
状态dp[p][i][j]可达之后，对于下一个点k：
dp[p|1<<k][j][k] = max{dp[p|1<<k][j][k], dp[p][i][j] + w[k] + w[k]*w[j] (+ W[三角形])}
P.S. 因为最好的汉密尔顿路径数目可能相当大，超出int所以使用long long来保存ways
代码如下：
/*
ID: j.sure.1
PROG:
LANG: C++
*/
/****************************************/
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <string>
#include <climits>
#include <iostream>
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
/****************************************/
int n, m;
const int N = 13;
bool G[N][N];
int w[N];
int dp[1<<N][N][N];
LL ways[1<<N][N][N];
void init()
{
memset(G, 0, sizeof(G));
memset(dp, -1, sizeof(dp));
memset(ways, 0, sizeof(ways));
}
void DP()
{
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) if(G[i][j]) {
dp[(1<<i)|(1<<j)][i][j] = w[i] + w[j] + w[i]*w[j];
ways[(1<<i)|(1<<j)][i][j] = 1;
}
}
for(int p = 0; p < (1<<n); p++) {
for(int i = 0; i < n; i++) if(p & (1<<i)) {
for(int j = 0; j < n; j++) if(p & (1<<j)) {
if(G[i][j] && dp[p][i][j] != -1) {
for(int k = 0; k < n; k++) {
if(G[j][k] && k != i && !(p & (1<<k))) {
int tmp = dp[p][i][j] + w[k] + w[k] * w[j];
if(G[i][k]) {
tmp += w[i] * w[j] * w[k];
}//形成三角形
if(dp[p|(1<<k)][j][k] < tmp) {
dp[p|(1<<k)][j][k] = tmp;
ways[p|(1<<k)][j][k] = ways[p][i][j];
}//决定是否更改原方案
else if(dp[p|(1<<k)][j][k] == tmp) {
ways[p|(1<<k)][j][k] += ways[p][i][j];
}
}
}
}
}
}
}
}
int main()
{
#ifdef J_Sure
// freopen("000.in", "r", stdin);
// freopen(".out", "w", stdout);
#endif
int T;
scanf("%d", &T);
while(T--) {
scanf("%d%d", &n, &m);
init();
for(int i = 0; i < n; i++) {
scanf("%d", &w[i]);
}
if(n == 1) {
printf("%d 1\n", w[0]);
continue ;
}
int u, v;
for(int i = 0; i < m; i++) {
scanf("%d%d", &u, &v);
u--; v--;
G[v][u] = G[u][v] = true;
}
DP();
int maxi = 0;
LL ans = 0;
int P = (1<<n) - 1;
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) if(G[i][j]) {
if(maxi < dp[P][i][j]) {
maxi = dp[P][i][j];
ans = ways[P][i][j];
}
else if(maxi == dp[P][i][j]) {
ans += ways[P][i][j];
}
}
}
printf("%d %lld\n", maxi, ans / 2);
}
return 0;
}