RQNOJ 35 营救 解题报告

　　SPFA，纯搜索，如果走到下一个位置的体力可以更少那就用那个更少的，如果一样多，看走的步数，用少的，就这样一个广搜。

　　代码如下：

#include 
#include 
#include 
#include 
#define QMAX 25000
struct node{
	int x, y;
}queue[QMAX];
int head, rear;
int map[500][500];
int used[500][500];

void enqueue(int x, int y)
{
	if(used[x][y]){
		return;
	}
	used[x][y] = 1;
	queue[rear].x = x;
	queue[rear].y = y;
	rear = (rear + 1) % QMAX;
}

void exqueue(int *x, int *y)
{
	*x = queue[head].x;
	*y = queue[head].y;
	used[*x][*y] = 0;
	head = (head + 1) % QMAX;
}

int dis[500][500];
int step[500][500];
int ans1 = 0x7FFFFF, ans2;

int main(int argc, char **argv)
{
	char c;
	int i, j;
	int m, n;
	int x, y;
	scanf("%d%d", &n, &m);
	scanf("%d%d", &x, &y);
	memset(dis, 0x7F, sizeof(dis));
	memset(step, 0x7F, sizeof(step));
	dis[x - 1][y - 1] = 0;
	step[x - 1][y - 1] = 1;
	enqueue(x - 1, y - 1);

	scanf("%d%d\n", &x, &y);
	for(i = 0; i < n; i++){
		for(j = 0; j < m; j++){
			scanf("%c", &c);
			map[i][j] = c - '0';
		}
		scanf("\n");
	}
	while(head != rear){
		exqueue(&i, &j);
#define deal(s, t) do{\
	if(s >= 0 && s < n && t >= 0 && t < m){\
		if((map[s][t] != 0) && (dis[s][t] >= dis[i][j] + abs(map[i][j] - map[s][t]))){\
			if(dis[s][t] > dis[i][j] + abs(map[i][j] - map[s][t])){\
				step[s][t] = step[i][j] + 1;\
				dis[s][t] = dis[i][j] + abs(map[i][j] - map[s][t]);\
				enqueue(s, t);\
			}else if(step[s][t] > step[i][j] + 1){\
				step[s][t] = step[i][j] + 1;\
				enqueue(s, t);\
			}\
		}\
	}\
}while(0)
		deal(i - 1, j - 1);
		deal(i - 1, j);
		deal(i - 1, j + 1);
		deal(i, j - 1);
		deal(i, j + 1);
		deal(i + 1, j - 1);
		deal(i + 1, j);
		deal(i + 1, j + 1);
	}
	if(dis[x - 1][y - 1] == 0x7F7F7F7F){
		printf("0 0\n");
	}else{
		printf("%d %d\n", step[x - 1][y - 1], dis[x - 1][y - 1]);
	}
	return 0;
}

　　

转载于:https://www.cnblogs.com/yylogo/archive/2011/09/25/RQNOJ-35.html