CF1252J Tiling Terrace

题目描述

Talia has just bought an abandoned house in the outskirt of Jakarta. The house has a nice and long yard which can be represented as a one-dimensional grid containing 1 \times N1×N cells. To beautify the house, Talia is going to build a terrace on the yard by tiling the cells. Each cell on the yard contains either soil (represented by the character '.') or rock (represented by the character '#'), and there are at most 5050 cells containing rocks.

Being a superstitious person, Talia wants to tile the terrace with mystical tiles that have the power to repel ghosts. There are three types of mystical tiles:

Type-1: Covers 1 \times 11×1 cell and can only be placed on a soil cell (".").
Type-2: Covers 1 \times 21×2 cells and can only be placed on two consecutive soil cells ("..").
Type-3: Covers 1 \times 31×3 cells and can only be placed on consecutive soil-rock-soil cells (".#.").

Each tile of Type-1, Type-2, and Type-3 has the power to repel G_1G1 , G_2G2 , and G_3G3 ghosts per day, respectively. There are also some mystical rules which must be followed for the power to be effective:

There should be no overlapping tiles, i.e. each cell is covered by at most one tile.
There should be at most KK tiles of Type-1, while there are no limitations for tiles of Type-2 and Type-3.

Talia is scared of ghosts, thus, the terrace (which is tiled by mystical tiles) should be able to repel as many ghosts as possible. Help Talia to find the maximum number of ghosts that can be repelled per day by the terrace. Note that Talia does not need to tile all the cells on the yard as long as the number of ghosts that can be repelled by the terrace is maximum.

输入格式

Input begins with a line containing five integers: NN KK G_1G1 G_2G2 G_3G3 ( 1 \le N \le 100,0001≤N≤100000 ; 0 \le K \le N0≤K≤N ; 0 \le G_1, G_2, G_3 \le 10000≤G1,G2,G3≤1000 ) representing the number of cells, the maximum number of tiles of Type-1, the number of ghosts repelled per day by a tile of Type-1, the number of ghosts repelled per day by a tile of Type-2, and the number of ghosts repelled by a tile of Type-3, respectively. The next line contains a string of NN characters representing the yard. Each character in the string is either '.' which represents a soil cell or '#' which represents a rock cell. There are at most 5050 rock cells.

输出格式

Output in a line an integer representing the maximum number of ghosts that can be repelled per day.

输入输出样例

输入 #1复制

输出 #1复制

输入 #2复制

输出 #2复制

输入 #3复制

输出 #3复制

说明/提示

Explanation for the sample input/output #1

Let "A" be a tile of Type-1, "BB" be a tile of Type-2, and "CCC" be a tile of Type-3. The tiling "ACCCBB" in this case produces the maximum number of ghosts that can be repelled, i.e. 10 + 40 + 25 = 7510+40+25=75

Explanation for the sample input/output #2

This sample input has the same yard with the previous sample input, but each tile of Type-2 can repel more ghosts per day. The tiling "BB#BBA" or "BB#ABB" produces the maximum number of ghosts that can be repelled, i.e. 100 + 100 + 10 = 210100+100+10=210 . Observe that the third cell is left untiled.

Explanation for the sample input/output #3

The tiling "ACCCA.#", "ACCC.A#", or ".CCCAA#" produces the maximum number of ghosts that can be repelled, i.e. 30 + 100 + 30 = 16030+100+30=160 . Observe that there is no way to tile the last cell.

题解：

非常容易判断是一道$DP$的题目。

联想一下这道题目：CF358D Dima and Hares

那么我们分方式决策，可以看出：第一种和第二种的选择都仅包含"."，只有第三种选择有"#"，那么我们可以考虑分段进行决策：将整个序列分成以"#"相隔开的段，并用$a[i]$数组保存$i$段中"."的个数。

那么我们设置状态$dp[i][j][0/1]$表示前$j$段用了$i$次方式$1$，后面的$0/1$表示最后的"#"有没有用方式$3$所获得的最大价值。

那么我们转移的时候就只需要考虑方式二的使用次数。

转移方程需要分类讨论。另外地，还有一种情况是不选择完这$n$个字符。这个东西可以在枚举方式1的时候用除以二自动取整来实现。

可以结合代码理解：

#include<cstdio>

#include<cstring>

#include<algorithm>

#define int long long

using namespace std;

const int maxn=1e5+10;

int n,k,g1,g2,g3,tot,ans;

char s[maxn];

int a[maxn],dp[60][maxn][5];

//dp[i][j][0/1]表示前j段使用i次方式1，最后一个"#"是否选用方式3所得的最大利益。

signed main()

{

    scanf("%I64d%I64d%I64d%I64d%I64d",&n,&k,&g1,&g2,&g3);

    scanf("%s",s+1);

    tot=1;

    for(int i=1;i<=n;i++)

    {

        if(s[i]=='#')

            tot++;

        else

            a[tot]++;

    }

    while(tot && !a[tot])

        tot--;

    if(!tot)

    {

        puts("0");

        return 0;

    }

    memset(dp,0xcf,sizeof(dp));

    dp[0][0][0]=0;

    for(int i=1;i<=tot;i++)

        for(int j=0;j<=k;j++)

        {

            int t=min(a[i],j);

            for(int l=0;l<=t;l++)

            {

                if(l<=a[i])

                    dp[i][j][0]=max(dp[i][j][0],dp[i-1][j-l][0]+(a[i]-l)/2*g2+l*g1);

                if(l<=a[i]-1)

                {

                    dp[i][j][0]=max(dp[i][j][0],dp[i-1][j-l][1]+(a[i]-l-1)/2*g2+g3+l*g1);

                    dp[i][j][1]=max(dp[i][j][1],dp[i-1][j-l][0]+(a[i]-l-1)/2*g2+l*g1);

                }

                if(l<=a[i]-2)

                    dp[i][j][1]=max(dp[i][j][1],dp[i-1][j-l][1]+(a[i]-l-2)/2*g2+g3+l*g1);

            }

        }

    for(int i=0;i<=k;i++)

        ans=max(ans,dp[tot][i][0]);

    printf("%I64d",ans);

    return 0;

}