題目連結：http://poj.org/problem?id=1179

Time Limit: 1000MS Memory Limit: 10000K

Description

Polygon is a game for one player that starts on a polygon with N vertices, like the one in Figure 1, where N=4. Each vertex is labelled with an integer and each edge is labelled with either the symbol + (addition) or the symbol * (product). The edges are numbered from 1 to N. 

On the first move, one of the edges is removed. Subsequent moves involve the following steps: 
pick an edge E and the two vertices V1 and V2 that are linked by E; and 
replace them by a new vertex, labelled with the result of performing the operation indicated in E on the labels of V1 and V2. 
The game ends when there are no more edges, and its score is the label of the single vertex remaining. 

Consider the polygon of Figure 1. The player started by removing edge 3. After that, the player picked edge 1, then edge 4, and, finally, edge 2. The score is 0. 

Write a program that, given a polygon, computes the highest possible score and lists all the edges that, if removed on the first move, can lead to a game with that score. 

Input

Your program is to read from standard input. The input describes a polygon with N vertices. It contains two lines. On the first line is the number N. The second line contains the labels of edges 1, ..., N, interleaved with the vertices' labels (first that of the vertex between edges 1 and 2, then that of the vertex between edges 2 and 3, and so on, until that of the vertex between edges N and 1), all separated by one space. An edge label is either the letter t (representing +) or the letter x (representing *). 

3 <= N <= 50 
For any sequence of moves, vertex labels are in the range [-32768,32767]. 

Output

Your program is to write to standard output. On the first line your program must write the highest score one can get for the input polygon. On the second line it must write the list of all edges that, if removed on the first move, can lead to a game with that score. Edges must be written in increasing order, separated by one space.

Sample Input

4
t -7 t 4 x 2 x 5

Sample Output

33
1 2

 

題意：

給出一個由無向邊和節點組成的環，每個節點上有一個數字，每條邊上有一個運算子（加或乘），

現在先割斷一條邊，然後環就成為一個鏈，然後你每次可以將這條鏈上的一條邊縮成一個點，產生的新點的權值就是兩個節點配合邊運算所產生的結果。

不停地縮邊成點，直到最後只有一個點為止，求這個點的權值最大是多少。

並給出所有能產生這個最大值的首先割斷的邊的編號，要求從小到大輸出。

 

題解：

區間DP，wyb出的毒瘤題，每次兩個小區間合併的時候，要記得有可能兩個最小的負數相乘可能會產生正數最大值。

因此需要同時維護區間最小值和最大值。

 

AC程式碼：

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
typedef pair<int,int> pii;
const int INF=0x3f3f3f3f;
const int maxn=55;

int n;
int op[2*maxn],nm[2*maxn];
pii dp[2*maxn][2*maxn];
inline int calc(int type,int a,int b){return type?a*b:a+b;}
inline void updatemn(int &x,int y){if(x>y) x=y;}
int solve(int l,int r)
{
    for(int s=2;s<=r-l+1;s++)
    {
        for(int st=l,ed=st+s-1;ed<=r;st++,ed++)
        {
            dp[st][ed].first=-INF;
            dp[st][ed].second=INF;
            for(int mid=st+1;mid<=ed;mid++)
            {
                pii le=dp[st][mid-1];
                pii ri=dp[mid][ed];

                int tmp1=calc(op[mid],le.first,ri.first);
                dp[st][ed].first=max(dp[st][ed].first,tmp1);
                dp[st][ed].second=min(dp[st][ed].second,tmp1);

                int tmp2=calc(op[mid],le.first,ri.second);
                dp[st][ed].first=max(dp[st][ed].first,tmp2);
                dp[st][ed].second=min(dp[st][ed].second,tmp2);

                int tmp3=calc(op[mid],le.second,ri.first);
                dp[st][ed].first=max(dp[st][ed].first,tmp3);
                dp[st][ed].second=min(dp[st][ed].second,tmp3);

                int tmp4=calc(op[mid],le.second,ri.second);
                dp[st][ed].first=max(dp[st][ed].first,tmp4);
                dp[st][ed].second=min(dp[st][ed].second,tmp4);
            }
        }
    }
    return dp[l][r].first;
}
int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        int m; char o[2];
        scanf("%s",o); op[i]=op[n+i]=(o[0]=='x');
        scanf("%d",&m); nm[i]=nm[n+i]=m;
    }

    for(int i=0;i<2*n;i++) dp[i][i]=make_pair(nm[i%n],nm[i%n]);
    int ans=-INF;
    for(int c=0;c<n;c++) ans=max(ans,solve(c,c+n-1));

    vector<int> E;
    for(int c=0;c<n;c++) if(dp[c][c+n-1].first==ans) E.push_back(c+1);
    sort(E.begin(),E.end());
    printf("%d\n",ans);
    for(int i=0;i<E.size();i++) printf("%d%c",E[i],(i==E.size()-1)?'\n':' ');
}

 

資料：

5
x 2 x 3 t 1 t 7 x 4

224
4


5
x -3 t -1 t -7 t -4 x -2

30
1 5


3
t 0 x 1 t -2

0
1


30
x 1 t 1 x 1 t 1 t 1 x 1 x 1 x 1 x 1 x 1 x 1 t 1 t 1 x 1 t 1 x 1 x 1 t 1 x 1 x 1 t 1 x 1 x 1 x 1 x 1 x 1 t 1 x 1 x 1 x 1

288
1 3 6 7 8 9 10 11 14 16 17 19 20 22 23 24 25 26 28 29 30


48
x 1 x 2 x 1 x -1 t 1 x -1 x -1 x 1 t 1 t -1 x 1 t 2 x 1 x 2 t 1 x 1 x -1 x -2 x 1 x 1 t 1 x 1 t 1 x 1 x 1 x 1 t 1 x 1 x 1 x 1 x 1 x 1 x 1 x -1 t 1 x 1 x -1 x -1 t 1 x 1 t 1 x 1 x 1 x -1 t 1 t -1 t -1 x 1

23328
45