zoj_2278-Fight for Food

//*************************************************************************************
//1.刚开始容易想到的DP状态是三维的即f[行][列][时间]。这样也可以求解，不过由于时间T可能会很大这样显然会超时。实际只要依次处理每一个会出
//现rat的点平且判断点与点之间是否能在要求的时间内到达就可以了，这样上只需要一维就可以了。
// 
//2.很容易想到按时间顺序对所有的点进行排序并且对于起点L不能到达的点要全部去掉。f[i]=max(f[j])+1 0<=j<i。f[i]表示在第i个点能够catch的最
//大数目的rat
//
//3.还有个重要的优化点,就是点与点之间的时间间隔有可能非常大。例如有的点rat在T=20s出现，有的点在T=1000s出现。其实可以对所有点按rat出现时
//间进行降序排列，并记录下从当前i点（位置[i,j]）到所有可到达点的最大距离maxDistance[i][j]。如果rat在点j出现的时间减去rat在点i出现的时间
//大于maxDistance[i][j]，那么显然小于j(0<=k<j)的所有点的rat从i点出发必然也都可以catch到
//*************************************************************************************

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define max(a,b) (((a) > (b)) ? (a) : (b))
#define N 10
#define NUM 30000

struct Rat{
	int row,col;
	int appear_time;
}rats[NUM];


bool bFlag[N][N];
int maxDistance[N][N];
int f[NUM];
int maxRatsNum[NUM];
int rowNum,colNum;
int ratsNum;

struct Position{
	int row,col;
	int distance;
};
int shorestDistance[N][N][N][N];
Position queue[N*N];

void getShorestDistance(int start_x,int start_y)
{
	if(shorestDistance[start_x][start_y][start_x][start_y]!=-1) return;

	int dir[][2]={{-1,0},{0,1},{1,0},{0,-1}};
	
	Position cur,next;
	int head=0,rear=0;
	cur.row=start_x,cur.col=start_y,cur.distance=0;
	queue[rear++] = cur;
	shorestDistance[start_x][start_y][start_x][start_y] = 0;

	while(head<rear)
	{
		cur = queue[head++];

		for(int i = 0;i<4;++i)
		{
			int newRow = cur.row+dir[i][0],newCol = cur.col+dir[i][1];
			if(newRow>=0&&newRow<rowNum&&newCol>=0&&newCol<colNum&&bFlag[newRow][newCol]&&shorestDistance[start_x][start_y][newRow][newCol]==-1)
			{
				next.row=newRow;
				next.col=newCol;
				next.distance = cur.distance+1;
				queue[rear++]=next;

				shorestDistance[start_x][start_y][newRow][newCol] = next.distance;
				maxDistance[start_x][start_y] = next.distance;
			}
		}
	}
}


int cmpRat(const void* a,const void* b)
{
	return (((Rat*)b)->appear_time-((Rat*)a)->appear_time);
}


int calc_max_rats()
{
	qsort(rats,ratsNum,sizeof(Rat),cmpRat);
	int row,col,newRow,newCol,leftTime;

	for(int i=0;i<ratsNum;++i)
	{
		row = rats[i].row,col=rats[i].col;
		f[i]=1;
		getShorestDistance(row,col);

		for(int j=i-1;j>=0;--j)
		{
			leftTime = rats[j].appear_time - rats[i].appear_time;
			if(leftTime>=maxDistance[row][col])
			{
				if(f[i]<maxRatsNum[j]+1) f[i] = maxRatsNum[j]+1;
				break;
			} 
			else
			{
				newRow = rats[j].row,newCol = rats[j].col;
				if(leftTime>=shorestDistance[row][col][newRow][newCol]&&f[i]<f[j]+1) f[i] = f[j]+1;
			}
		}

		maxRatsNum[i]=(i>0?max(maxRatsNum[i-1],f[i]):f[0]);
	}

	return ratsNum>0?maxRatsNum[ratsNum-1]:0;
}


int main()
{
	//freopen("input.txt","r",stdin);
	char ch;
	while(scanf("%d %d",&rowNum,&colNum)==2)
	{
		memset(bFlag,false,sizeof(bFlag));
		memset(shorestDistance,255,sizeof(shorestDistance));

		int start_x,start_y;
		for(int i = 0;i<rowNum;++i)
		{
			getchar();
			for(int j = 0;j<colNum;++j)
			{
				scanf("%c",&ch);
				if(ch=='.') bFlag[i][j] = true;
				else if(ch=='L'){
					start_x=i;
					start_y=j;
					bFlag[i][j] = true;
				}
			}
		}

		getShorestDistance(start_x,start_y);

		scanf("%d",&ratsNum);
		int row,col,appear_time;
		int i=0,distance;

		while(ratsNum>0)
		{
			scanf("%d %d %d",&row,&col,&appear_time);
			distance = shorestDistance[start_x][start_y][--row][--col];
			
			if(distance!=-1&&appear_time>=distance){
				rats[i].row = row;
				rats[i].col = col;
				rats[i].appear_time = appear_time;
				++i;
			}
			--ratsNum;
		}

		ratsNum = i;
		printf("%d\n",calc_max_rats());
	}

	return 0;
}