[IOI1998]Polygon

题意翻译

题目可能有些许修改，但大意一致

多边形是一个玩家在一个有n个顶点的多边形上的游戏，如图所示，其中n＝4。每个顶点用整数标记，每个边用符号+（加）或符号*（乘积）标记。

第一步，删除其中一条边。随后每一步：

选择一条边连接的两个顶点V1和V2，用边上的运算符计算V1和V2得到的结果来替换这两个顶点。

游戏结束时，只有一个顶点，没有多余的边。

如图所示，玩家先移除编号为3的边。之后，玩家选择计算编号为1的边，然后计算编号为4的边，最后，计算编号为2的边。结果是0。

（翻译者友情提示：这里每条边的运算符旁边的数字为边的编号，不拿来计算）

编写一个程序，给定一个多边形，计算最高可能的分数。

输入格式

输入描述一个有n个顶点的多边形，它包含两行。第一行是数字n，为总边数。

第二行描述这个多边形，一共有2n个读入，每两个读入中第一个是字符，第二个是数字。

第一个字符为第一条边的计算符号(t代表相加，x代表相乘)，第二个代表顶点上的数字。首尾相连。

3 < = n < = 50

对于任何一系列的操作，顶点数字都在[-32768,32767]的范围内。

输出格式

第一行，输出最高的分数。在第二行，它必须写出所有可能的被清除后的边仍能得到最高得分的列表，必须严格递增。

题目描述

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.

输入输出格式

输入格式：

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].

输出格式：

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.

输入输出样例

输入样例#1：

4

t -7 t 4 x 2 x 5

输出样例#1：

33

1 2

题解

• 在做这道题目之前,相信大家之前已经做过这道题目,我们发现在枚举删掉哪条边后,就和石子合并非常相似,那我们可不可以以相似的思路来做这道题目呢?
• 如果按照上一题的思路,我们需要定义一个dp数组,其中dp[l,r]表示从l到r内的最大值.但是这样定义状态是有误的.dp[l,r]无法从dp[l,k]+dp[k+1,r]转移而来,因为[l,k]内的最小值与[k+1,r]内的最小值可能是负数,两者相乘,结果可能比两个区间内的最大值相乘更大.
• 那么我们想,可不可以保存一段区间内最小值和最大值两个关键值呢?答案是肯定的.因为最大值的来源只可能是两个区间内最大值相加,相乘或者是两个区间内的最小值相乘,而最小值只可能是两个区间内的最小值相加,或者是一个最大值去和另一个最小值相乘.
• 我们定义dp[l][r][0]表示[l,r]区间内的最大值,dp[l][r][1]表示[l,r]内的最小值

• 那么就很容易得出状态转移方程

  若op为' + '

  \[dp[l,r][0]=max(dp[l,r][0],dp[l,k][0]+dp[k+1,r][1])\]

  \[dp[l,r][1]=min(dp[l,r][1],dp[l,k][1]+dp[k+1,r][1])\]

  若op为' * '

  \[dp[l,r][0]=max(dp[l,r][0],max(dp[l,k][0]*dp[k+1,r][0],dp[l,k][1]*dp[k+1,r][1]))\]

  \[dp[l,r][1]=min(dp[l,r][1],min(dp[l,k][0]*dp[k+1,r][1],dp[l,k][1]*dp[k+1,r][0]))\]

• 事实上,我们其实还可以省去枚举删除起始边的那一重循环,试想一下,无论我们从哪个地方断开边后我们都要将原数列复制一遍接在最后,形成一个2*N长的新数列,这样的话我们可以默认从1处断开,然后就是标准的区间dp\(O(n^3)\)枚举

  最后的答案就是\(max(dp[i][i+n-1][0])_{1<=i<=n}\)

#include
    
     #define MIN(a,b) (a)>(b)?(b):(a)#define MAX(a,b) (a)>(b)?(a):(b)#define in(i) (i=read())using namespace std;int read(){    int ans=0,f=1;    char i=getchar();    while(i<'0'||i>'9'){        if(i=='-') f=-1;        i=getchar();    }    while(i>='0' && i<='9'){        ans=(ans<<1)+(ans<<3)+i-'0';        i=getchar();    }    return ans*f;}const int inf=20000000;int n,cnt;int dp[150][150][2];int a[110];char op[110];inline void init(){    for(int i=1;i<=2*n;i++){        for(int j=i;j<=2*n;j++){            dp[i][j][0]=-inf,dp[i][j][1]=inf;        }    }    for(int i=1;i<=n;i++){        dp[i][i][0]=dp[i][i][1]=dp[i+n][i+n][0]=dp[i+n][i+n][1]=a[i];    }}void dp1(int l,int r,int k){    dp[l][r][0]=MAX(dp[l][r][0],dp[l][k][0]+dp[k+1][r][0]);    dp[l][r][1]=MIN(dp[l][r][1],dp[l][k][1]+dp[k+1][r][1]);}void dp2(int l,int r,int k){    dp[l][r][0]=MAX(dp[l][r][0],MAX(dp[l][k][0]*dp[k+1][r][0],dp[l][k][1]*dp[k+1][r][1]));    dp[l][r][1]=MIN(dp[l][r][1],MIN(dp[l][k][0]*dp[k+1][r][1],dp[l][k][1]*dp[k+1][r][0]));}int main(){    int ans=-inf;    in(n);    for(int i=1;i<=n;i++){        cin>>op[i];in(a[i]);        op[i+n]=op[i];a[i+n]=a[i];    }    init();    for(int len=2;len<=n;len++){        for(int l=1;l<=2*n;l++){            int r=l+len-1;            for(int k=l;k