G - Islands and Bridges解题报告（来自网络）

G - Islands and Bridges

Time Limit:4000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  POJ 2288

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

题目描述：哈密尔顿路问题。n个点，每一个点有权值，设哈密尔顿路为 C1C2...Cn，Ci的权值为Vi，一条哈密尔顿路的值分为三部分计算：
1.每一个点的权值之和
2.对于图中的每一条CiCi+1，加上Vi*Vi+1
3.对于路径中的连续三个点：CiCi+1Ci+2，若在图中，三点构成三角形，则要加上Vi*Vi+1*Vi+2
求一条汉密尔顿路可以获得的最大值，并且还要输出有多少条这样的哈密尔顿路。
 
这道题的状态感觉不是很难想，因为根据一般的哈密尔顿路问题，首先想到的是设计二维状态，dp[i , s]表示当前在i点，走过的点形成状态集合s。但是这道题在求解值的时候有一个不一样的地方，就是第三部分，如果还是设计成二维的状态，就会很麻烦，因为每加入一个新点，要判断新点、当前点、倒数第二个点是否构成三角形，所以要记录倒数第二个点。很自然地想到扩展状态的维数，增加一维，记录倒数第二个点。
1>　　设计状态：
dp[i , j , s]表示当前站在j点，前一个点是i点，形成的状态集合是s，此时的最大值，way[i , j , s]记录当前状态下达到最大值的路径数；
2>　　状态转移：
设k点不在集合s中，且存在边<j , k>
设q为下步到达k点获得的最大值
令r = s + （1<<k），为当前站在点k，前一个点为j，形成状态集合r
若i，j，k形成三角形，则q = dp[i][j][s] + v[k] + v[j]*v[k] + v[i]*v[j]*v[k];
否则，q = dp[i][j][s] + v[k] + v[j]*v[k];
若q大于dp[j][k][r]；则：
dp[j][k][r] = q
way[j][k][r] = way[i][j][s];
若q等于dp[j][k][r]，则：
way[j][k][r] += way[i][j][s];
3>　　初始化：
显然，若i点到j点有边，则： 
dp[i][j][(1<<i)+(1<<j)] = v[i] + v[j] + v[i]*v[j];
way[i][j][(1<<i)+(1<<j)] = 1;
 4>　　结果的产生：
最后的结果我们要枚举点i和j，找到最大的dp[i][j][(1<<n)-1]，并且更新记录路径数ansp，最后ansp要除2才是结果，因为题目最后一句话，正向反向是一样的路。
 
此外，需要注意的是discuss提到的特殊情况，要用__int64，并且注意n等于1时，最大值就是第一个点的权值，路径数为1。
（ps：在处理特殊情况时，忘记换行，还PE一次，这年头PE的还真少见啊。。。）
代码如下，祝1Y。

#include <stdio.h>
 #include <string.h>
 
 typedef __int64 i64;
 
 const int maxn = 13;
 const int maxs = 1<<maxn | 1;
 
 i64 dp[maxn][maxn][maxs], way[maxn][maxn][maxs];
 int map[maxn][maxn], v[maxn];
 int n, m, S;
 
 void StateCompressDp() {
 
     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 (map[i][j]) {
                 dp[i][j][(1<<i)+(1<<j)] = v[i] + v[j] + v[i]*v[j];
                 way[i][j][(1<<i)+(1<<j)] = 1;
             }
 
         }
     }
     for (int p = 3; p < S; p++) {
         for (int i = 0; i < n; i++) {
             if (!(p&1<<i)) continue;
             for (int j = 0; j < n; j++) {
                 if (i == j || !(p&1<<j) || dp[i][j][p] == -1) continue;
                 for (int k = 0; k < n; k++) {
                     if (p&1<<k || !map[j][k]) continue;
                     int r = p + (1 << k);
                     i64 q = dp[i][j][p] + v[k] + v[j]*v[k];
                     if (map[i][k]) {
                         q += v[i] * v[j] * v[k];
                     }
                     if (q > dp[j][k][r]) {
                         dp[j][k][r] = q;
                         way[j][k][r] = way[i][j][p];
                     } else if (q == dp[j][k][r]) {
                         way[j][k][r] += way[i][j][p];
                     }
                 }
             }
         }
     }
 
     return ;
 }
 
 int main () {
 
     int t;
 
     scanf("%d", &t);
     while (t--) {
         int tmpx, tmpy;
         memset(map, 0, sizeof(map));
         scanf("%d%d", &n, &m);
         for (int i = 0; i < n; i++) {
             scanf("%d", &v[i]);
         }
         for (int i = 0; i < m; i++) {
             scanf("%d%d", &tmpx, &tmpy);
             map[tmpx-1][tmpy-1] = map[tmpy-1][tmpx-1] = 1;
         }
         S = 1 << n;
         if (n == 1) {
             printf("%d %d\n", v[0], 1);
         } else {
             StateCompressDp();
             i64 ansv = -1, ansp = 0;
             for (int i = 0; i < n; i++) {
                 for (int j = 0; j < n; j++) {
                     if (i == j) continue;
                     if (dp[i][j][S-1] > ansv) {
                         ansv = dp[i][j][S-1];
                         ansp = way[i][j][S-1];
                     } else if (dp[i][j][S-1] == ansv) {
                         ansp += way[i][j][S-1];
                     }
                 }
             }
             printf("%I64d %I64d\n", ansv == -1 ? 0 : ansv, ansp/2);
         }
     }
 
     return 0;
 }