Code Jam 2017 Qualification Round Problem D. Fashion Show

题意

You are about to host a fashion show to show off three new styles of clothing. The show will be held on a stage which is in the most fashionable of all shapes: an N-by-N grid of cells.

Each cell in the grid can be empty (which we represent with a . character) or can contain one fashion model. The models come in three types, depending on the clothing style they are wearing: +, x, and the super-trendy o. A cell with a + or x model in it adds 1 *style point*to the show. A cell with an o model in it adds 2 style points. Empty cells add no style points.

To achieve the maximum artistic effect, there are rules on how models can be placed relative to each other.

• Whenever any two models share a row or column, at least one of the two must be a +.
• Whenever any two models share a diagonal of the grid, at least one of the two must be an x.

Formally, a model located in row $i_0$ and column $j_0$ and a model located in row $i_1$ and column $j_1$ share a row if and only if $i_0$ = $i_1$, they share a column if and only if $j_0$ = $j_1$, and they share a diagonal if and only if $i_0$ + $j_0$ = $i_1$ + $j_1$ or $i_0$ - $j_0$ = $i_1$ - $j_1$.

For example, the following grid is not legal:

...
x+o
.+.

The middle row has a pair of models (x and o) that does not include a +. The diagonal starting at the + in the bottom row and running up to the o in the middle row has two models, and neither of them is an x.

However, the following grid is legal. No row, column, or diagonal violates the rules.

+.x
+x+
o..

Your artistic advisor has already placed M models in certain cells, following these rules. You are free to place any number (including zero) of additional models of whichever types you like. You may not remove existing models, but you may upgrade as many existing +and x models into o models as you wish, as long as the above rules are not violated.

Your task is to find a legal way of placing and/or upgrading models that earns the maximum possible number of style points.

解法

首先，整理一下题意，可以看到：

对于 N * N 的二维图。每行（列）最多有一个 x 或 o 。每条斜线上最多有一个 + 或 o 。要求在二维图中放入 x 、+ 、o ，使得最终所得的分数最大，其中 x 和 + 计 1 分， o 计 2 分。

简单思考可以得出，o 可以视作是 x 和 + 在同一格的叠加，且由于行列与斜线之间的关系无相互影响，故可将该问题视作两个子问题：

1. 在二维图中尽可能多的放入 x 使得分数最大。
2. 在二维图中尽可能多的放入 + 使得分数最大。

最后将两个图进行叠加即可处理出所需输出的答案。

子问题 1：

每行（列）最多放一个 x ，使得 x 尽可能地多。这个子问题比较简单，直接贪心可解，且放置 x 的个数一定是 N 个。

子问题 2：

每条斜线上最多出现一个 + ，使得 + 尽可能多。

首先将图进行 45 度旋转，将其仍视作是同行（列）不能有多个 +。

其中，在原二维图不存在 + 时，最多可摆放的 + 为 $2\times N - 2$ 个。将最外层视作是 N 层 ，次外层为 N-1 层。则当 N-1 层存在一个 + 将使得最外层失去两个 + ，同理，每层与其内层的关系均如此。故为尽可能多的摆放，则应从最外层开始，螺旋状地想内层枚举，若存在点符合摆放条件，则即时放入 + 。

代码

#include<bits/stdc++.h>
using namespace std;
char ori[110][110], chg[110][110];
bool rock[110][110], bishop[110][110];
bool mrkR[110], mrkC[110];
set<int> lft, rgt;
int r, c, n, m;
char C;
void init()
{
    memset(rock, 0, sizeof(rock));
    memset(bishop, 0, sizeof(bishop));
    memset(mrkR, 0, sizeof(mrkR));
    memset(mrkC, 0, sizeof(mrkC));
    memset(ori, 0, sizeof(ori));
    memset(chg, 0, sizeof(chg));

    lft.clear();    rgt.clear();
}
int main()
{
    freopen("G:\\D.in", "r", stdin);
    freopen("G:\\D.out", "w", stdout);
    int T;
    scanf("%d",&T);
    for(int ica=1;ica<=T;ica++)
    {
        scanf("%d %d",&n,&m);
        init();
        for(int i=0;i<m;i++)
        {
            scanf(" %c %d %d",&C, &r,&c);
            ori[r][c] = C;
            if(C != '+')
                rock[r][c] = 1, mrkR[r] = 1, mrkC[c] = 1;
            if(C != 'x')
                bishop[r][c] = 1,   lft.insert(r-c),   rgt.insert(r+c);
        }

        // solve rock
        r = 1,  c = 1;
        for(;r<=n;r++)
        {
            if(mrkR[r]) continue;
            while(mrkC[c])  c++;
            rock[r][c] = 1;
            mrkR[r] = 1,    mrkC[c] = 1;
        }

        //solve bishop
        for(int i=1;i<=n;i++)
        {
            // row = i
            r = i;
            for(c=i;c<=n-i+1;c++)
                if(lft.find(r-c) == lft.end() && rgt.find(r+c) == rgt.end())
                    bishop[r][c] = 1,   lft.insert(r-c),    rgt.insert(r+c);

            //row=n-i
            r = n-i+1;
            for(c=i;c<=n-i+1;c++)
                if(lft.find(r-c) == lft.end() && rgt.find(r+c) == rgt.end())
                    bishop[r][c] = 1,   lft.insert(r-c),    rgt.insert(r+c);

            // col = i
            c = i;
            for(r=i;r<=n-i+1;r++)
                if(lft.find(r-c) == lft.end() && rgt.find(r+c) == rgt.end())
                    bishop[r][c] = 1,   lft.insert(r-c),    rgt.insert(r+c);

            //col = n-i
            c = n-i+1;
            for(r=i;r<=n-i+1;r++)
                if(lft.find(r-c) == lft.end() && rgt.find(r+c) == rgt.end())
                    bishop[r][c] = 1,   lft.insert(r-c),    rgt.insert(r+c);
        }

        int y = 0;
        vector< pair<int, int> > ans;
        ans.clear();
        for(r=1;r<=n;r++)
        {
            for(c=1;c<=n;c++)
            {
                y += rock[r][c] + bishop[r][c];
                if(rock[r][c] && bishop[r][c])
                    chg[r][c] = 'o';
                else if(rock[r][c])
                    chg[r][c] = 'x';
                else if(bishop[r][c])
                    chg[r][c] = '+';
                if(chg[r][c] != ori[r][c])
                    ans.push_back(make_pair(r, c));
            }
        }
        printf("Case #%d: %d %d\n", ica, y, ans.size());
        for(int i=0;i<ans.size();i++)
        {
            r = ans[i].first;   c = ans[i].second;
            printf("%c %d %d\n", chg[r][c], r, c);
        }
    }
}