UVA-11882 bfs + dfs + 剪枝

        假设当前已经到达(x,y)，用bfs判断一下还可以到达的点有maxd个，如果maxd加上当前已经经过的长度小于当前答案的长度就退出，如果相同，就将bfs搜索到的点从大到小排序，如果连最大序列都无法大于当前答案，就直接退出，否则继续搜索。

       但是这题时间要求极高，不可以同时搜索同时更新答案，必须直到完成一种搜索（到达边界）才能更新答案，否则必定超时。

AC代码：

#include
#include
#include
#include
using namespace std;

const int maxn = 30 + 5;
char G[maxn][maxn];
int vis[maxn][maxn], d[maxn][maxn];
int n, m;

int ans[maxn], len; //答案以及答案的长度
int a[maxn];

struct node{
	int x, y;
	node(){
	}
	node(int x, int y):x(x), y(y){
	}
};

const int dx[] = {1,-1,0,0};
const int dy[] = {0,0,1,-1};



int bfs(int x, int y, int cur){ //从cur开始保存 
	int c = cur;
	memcpy(d, vis, sizeof(vis));
	queueq;
	q.push(node(x, y));
	d[x][y] = 1;
	while(!q.empty()){
		node pos = q.front();
		q.pop();
		x = pos.x, y = pos.y;
		for(int i = 0; i < 4; ++i){
			int px = x + dx[i], py = y + dy[i];
			if(px < 0 || px >= n || py < 0 || py >= m) continue;
			if(G[px][py] == '#' || d[px][py]) continue;
			d[px][py] = 1;
			a[c++] = G[px][py] - '0';
			q.push(node(px, py));
		}
	}	
	return c;
}

bool cmp(int a, int b){
	return a > b;
}

void dfs(int x, int y, int cur){
	int h = bfs(x, y, cur);
	
	if(h == cur) {  //边界 
		int flag = 0;
		if(cur < len) return;
		else if(cur > len) flag = 1;
		else for(int i = 0; i < len; ++i){
			if(a[i] < ans[i]) return;
			else if(a[i] > ans[i]) {
				flag = 1;
				break;
			}
		}
		if(flag) {
			len = cur;
			for(int i = 0; i < len; ++i) ans[i] = a[i]; //update
		}
		return;
	}
	
	if(h < len) return;
	else if(h == len){
		sort(a + cur, a + h, cmp);
		int flag = 0;
		for(int i = 0; i < len; ++i) {
			if(a[i] < ans[i]) return;
			else if(a[i] > ans[i]) {
				flag = 1;
				break;
			}
		}
		if(!flag) return;
	}

	
	for(int i = 0; i < 4; ++i){
		int px = x + dx[i], py = y + dy[i];
		if(px < 0 || px >= n || py < 0 || py >= m) continue;
		if(G[px][py] == '#' || vis[px][py]) continue;
		a[cur] = G[px][py] - '0';
		vis[px][py] = 1;
		dfs(px, py, cur + 1);
		vis[px][py] = 0;
	}
}


int  main(){
	while(scanf("%d%d", &n, &m) == 2 && n){
		memset(ans, 0, sizeof(ans));
		len = 0;
		for(int i = 0; i < n; ++i) scanf("%s", G[i]);
		
		for(int i = 0; i < n; ++i)
		for(int j = 0; j < m; ++j) {
			if(G[i][j] == '#') continue;
			memset(vis, 0, sizeof(vis));
			a[0] = G[i][j] - '0';
			vis[i][j] = 1;
			dfs(i, j, 1);
			vis[i][j] = 0;
		}
		
		for(int i = 0; i < len; ++i){
			printf("%d", ans[i]);
		}
		printf("\n");
	}
	return 0;
}

还有一种dfs代码，思路同上

#include
#include
#include
#include
using namespace std;

const int maxn = 30 + 5;
char G[maxn][maxn];
int vis[maxn][maxn], d[maxn][maxn];
int n, m;

int ans[maxn], len; //答案以及答案的长度
int a[maxn];

struct node{
	int x, y;
	node(){
	}
	node(int x, int y):x(x), y(y){
	}
};

const int dx[] = {1,-1,0,0};
const int dy[] = {0,0,1,-1};

bool cmp(int a, int b){
	return a > b;
}

int bfs(int x, int y, int cur){
	int c = cur;
	memcpy(d, vis, sizeof(vis));
	queueq;
	q.push(node(x, y));
	d[x][y] = 1;
	while(!q.empty()){
		node pos = q.front();
		q.pop();
		x = pos.x, y = pos.y;
		for(int i = 0; i < 4; ++i){
			int px = x + dx[i], py = y + dy[i];
			if(px < 0 || px >= n || py < 0 || py >= m) continue;
			if(G[px][py] == '#' || d[px][py]) continue;
			d[px][py] = 1;
			a[c++] = G[px][py] - '0';
			q.push(node(px, py));
		}
	}	
	return c;
}


void dfs(int x, int y, int cur){
	a[cur++] = G[x][y] - '0'; 
	int h = bfs(x, y, cur);
	
	if(h == cur) {  //边界 
		int flag = 0;
		if(cur < len) return;
		else if(cur > len) flag = 1;
		else for(int i = 0; i < len; ++i){
			if(a[i] < ans[i]) return;
			else if(a[i] > ans[i]) {
				flag = 1;
				break;
			}
		}
		if(flag) {
			len = cur;
			for(int i = 0; i < len; ++i) ans[i] = a[i]; //update
		}
		return;
	}
	
	if(h < len) return;
	else if(h == len){
		sort(a + cur, a + h, cmp);
		int flag = 0;
		for(int i = 0; i < len; ++i) {
			if(a[i] < ans[i]) return;
			else if(a[i] > ans[i]) {
				flag = 1;
				break;
			}
		}
		if(!flag) return;
	}

	vis[x][y] = 1;
	for(int i = 0; i < 4; ++i){
		int px = x + dx[i], py = y + dy[i];
		if(px < 0 || px >= n || py < 0 || py >= m) continue;
		if(G[px][py] == '#' || vis[px][py]) continue;
		dfs(px, py, cur);
	}
	vis[x][y] = 0;
	
}

int  main(){
	while(scanf("%d%d", &n, &m) == 2 && n){
		memset(ans, 0, sizeof(ans));
		len = 0;
		for(int i = 0; i < n; ++i) scanf("%s", G[i]);
		
		
		for(int i = 0; i < n; ++i)
		for(int j = 0; j < m; ++j) {
			if(G[i][j] == '#') continue;
			memset(vis, 0, sizeof(vis));
			dfs(i, j, 0);
		}
		for(int i = 0; i < len; ++i){
			printf("%d", ans[i]);
		}
		printf("\n");
	}
	return 0;
}

如有不当之处欢迎指出！

转载于:https://www.cnblogs.com/flyawayl/p/8305448.html