POJ 1826 The Best Farm
| Time Limit: 3000MS | Memory Limit: 30000K | |
| Total Submissions: 1519 | Accepted: 455 |
Description
In a fight against invaders, farmer William get the farmers all around the country together to help the king beat the invaders. So the king decided to award farmer William a great farm after the winning of the war.
Problem
The king divides his country as a A*B grid and labels each 1*1 square with a pair of integers. Refer to the following picture as example:
Nevertheless, not all the squares are available. Some have been awarded to others, and some have been destroyed in the war. So the king only list the available 1*1 squares, and let farm William choose some from them. Meanwhile, William can’t choose all of the squares. He can only choose some of them, so that the squares can form a connected area, to build up his farm. A connected area is defined as following:
1. A connected area is made up of some 1*1 squares;
2. From any of these 1*1 squares, one can walk to any other 1*1 square belonged to this area without entering some square that is not in this area;
3. When standing in a square, one can walk into the adjacent squares in four directions: north, south, east, and west.
In addition, every square available has a value. William should choose to build up a farm (connected area) that is the most valuable. In other words, William should choose some of the square forming a connected area, in which the sum of the value of squares is the largest.
Your task in this problem is to find out the largest value.
Input
Output
Sample Input
1 0 0 1 6 0 1 1 0 0 1 1 0 1 2 2 2 2 1 2 2 -1 1 0
Sample Output
1 4
Source
Atlas of rruucc@POJ
/* 主要思想就是哈希，因为坐标值x,y非常大，如果不哈希是无法存储的剩下的就是 找权值最大的连通分量了，一开始没有看清题，以为每个farm的value值可以是负 数(如果是这样就需要树状DP了),写着写着突然发现题目中说value值是非负值， 那么这就容易多了，直接DFS找所有连通分量然后取总value最大的就是题目要求 的了。看来好看清题啊，不过乘这个机会又把树状DP复习了一下 */ #include <iostream> //定义哈希List的最大容量 #define HASH_VAL 200000 using namespace std; //哈希容器中元素的定义 struct elem { //x, y是当前地的坐标, pos是当前地在farms数组中对应的下标 int x, y, pos; //指向下一个具有同样哈希值的elem，避免访问冲突 elem *next; elem() { x = y = pos = 0; next = NULL; } }; //哈希容器 struct hash { //next指示第一个具有这个哈希值的farm elem *next; }hashs[HASH_VAL + 2]; //存储每一块farm struct farm { int value; //标识是否被访问过 bool v; elem *ptr; }farms[HASH_VAL + 2]; int fNum, maxVal; //哈希函数 int getHashVal(int x, int y) { __int64 temp = x * x + y * y; return temp % HASH_VAL; } //-1 is not exist, else the pos of the farms in farms //if not exist and insertIfNotExist is true then insert this farm to the hashs int getPos(int x, int y, bool insertIfNotExist, int pos, int value) { int hashVal = getHashVal(x, y); if(!hashs[hashVal].next) { if(!insertIfNotExist) return -1; else { elem *newPtr = new elem(); newPtr->x = x; newPtr->y = y; newPtr->pos = pos; farms[pos].ptr = newPtr; farms[pos].v = false; farms[pos].value = value; hashs[hashVal].next = newPtr; return pos; } } if(hashs[hashVal].next->x == x && hashs[hashVal].next->y == y) return hashs[hashVal].next->pos; elem *curPtr = hashs[hashVal].next; while(curPtr->next) { if(curPtr->next->x == x && curPtr->next->y == y) return curPtr->next->pos; curPtr = curPtr->next; } if(!insertIfNotExist) return -1; else { elem *newPtr = new elem(); newPtr->x = x; newPtr->y = y; newPtr->pos = pos; farms[pos].ptr = newPtr; farms[pos].v = false; farms[pos].value = value; curPtr->next = newPtr; return pos; } } int dir[4][2] = {{0, 1}, {0, -1}, {-1, 0}, {1, 0}}; //dfs遍历找连通分量 int dfs(int nodePos) { farms[nodePos].v = true; int curVal = farms[nodePos].value; int curX = farms[nodePos].ptr->x, curY = farms[nodePos].ptr->y; int newX, newY; for(int p = 0; p < 4; p++) { newX = curX + dir[p][0]; newY = curY + dir[p][1]; int pos = getPos(newX, newY, false, 0, 0); if(pos == -1 || farms[pos].v) continue; curVal += dfs(pos); } return curVal; } int main() { int i, x, y, value; while(scanf("%d", &fNum) && fNum != 0) { maxVal = INT_MIN; memset(hashs, 0, sizeof(hashs)); for(i = 0; i < fNum; i++) { scanf("%d%d%d", &x, &y, &value); if(farms[i].ptr != NULL) { delete farms[i].ptr; farms[i].ptr = NULL; } getPos(x, y, true, i, value); } int curVal; for(i = 0; i < fNum; i++) { if(farms[i].v) continue; curVal = dfs(i); if(curVal > maxVal) maxVal = curVal; } printf("%d/n", maxVal); } return 0; }