2014
01-04

# Continuous Same Game (1)

Continuous Same Game is a simple game played on a grid of colored blocks. Groups of two or more connected (orthogonally, not diagonally) blocks that are the same color may be removed from the board. When a group of blocks is removed, the blocks above those removed ones fall down into the empty space. When an entire column of blocks is removed, all the columns to the right of that column shift to the left to fill the empty columns. Points are scored whenever a group of blocks is removed. The number of points per block increases as the group becomes bigger. When N blocks are removed, N*(N-1) points are scored.

LL was interested in this game at one time, but he found it is so difficult to find the optimal scheme. So he always play the game with a greedy strategy: choose the largest group to remove and if there are more than one largest group with equal number of blocks, choose the one which contains the most preceding block ( (x1,y1) is in front of (x2,y2) if and only if (x1<x2 || x1==x2 && y1<y2) ). Now, he want to know how many points he will get. Can you help him?

Each test case begins with two integers n,m ( 5<=n,m<=20 ), which is the size of the board. Then n lines follow, each contains m characters, indicating the color of the block. There are 5 colors, and each with equal probability.

Each test case begins with two integers n,m ( 5<=n,m<=20 ), which is the size of the board. Then n lines follow, each contains m characters, indicating the color of the block. There are 5 colors, and each with equal probability.

5 5
35552
31154
33222
21134
12314

32

Hint
35552    00552    00002    00002    00000    00000
31154    05154    05104    00004    00002    00000
33222    01222    01222    00122    00104    00100
21134    21134    21134    25234    25234    25230
12314    12314    12314    12314    12314    12312

The total point is 12+6+6+2+6=32.


http://acm.hit.edu.cn/hoj/problem/view?id=2258

#include <stdio.h>
#include <math.h>

const double pi = acos(-1);

int main()
{
int t, n, m, i, c;
double angle, r, arclength[1024], circle, totalength;

scanf("%d",&t);
while (t--)
{
scanf("%d %d", &n, &m);

angle = 2*pi/n;
c = m / n;
m = m % n;
totalength = 0, circle = 0;

for (i = 1; i <= n / 2; i++)
{
r = sqrt(2 - 2 * cos(i*angle) );
arclength[i] = arclength[n-i] = angle * r;
}
arclength[n] = 0;

for (i = 1; i <= n; i++)
circle += arclength[i];
for (i = 1; i <= m; i++)
totalength += arclength[i];

totalength += c*circle;
printf("%.2lf\n", totalength);
}

return 0;
}

1. 思路二可以用一个长度为k的队列来实现，入队后判断下队尾元素的next指针是否为空，若为空，则出队指针即为所求。

2. 问题3是不是应该为1/4 .因为截取的三段，无论是否能组成三角形， x， y-x ，1-y,都应大于0，所以 x<y,基础应该是一个大三角形。小三角是大三角的 1/4.