2014 ACM/ICPC Asia Regional 广州网赛 | HDU 5023 & HDU 5024 & HDU 5025

Doubi Gao3 posted @ 2014年9月23日 22:25 in acm with tags icpc , 7193 阅读

HDU 5203：给定一段区间，初始颜色为2，你可以将一整段区间涂成某一种颜色，询问一段区间上的颜色数。

线段树操作。由于颜色最多只有30种，我们可以利用一个int来处理所有的颜色，向下用&运算，向上用|运算来合并所有颜色即可。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <map>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 1000100;

#define lc o<<1
#define rc o<<1|1

struct SegmentTree
{
    int color[maxn<<2];
    int setv[maxn<<2];
    
    int n;
    
    void init(int n)
    {
        this->n = n;
    }
    
    void pushup(int o)
    {
        color[o] = color[lc] | color[rc];
    }
    
    void build(int o, int L, int R)
    {
        color[o] = (1<<1); setv[o] = 2;
        if(L == R) return ;
        int M = L+(R-L)/2;
        build(lc, L, M);
        build(rc, M+1, R);
        pushup(o);
    }
    
    void pushdown(int o, int L, int R)
    {
        if(setv[o])
        {
            color[lc] = (1<<(setv[o]-1));
            color[rc] = (1<<(setv[o]-1));
            
            setv[lc] = setv[rc] = setv[o];
            setv[o] = 0;
        }
    }
    
    void update(int o, int L, int R, int y1, int y2, int v)
    {
        if(y1 <= L && y2 >= R)
        {
            color[o] = (1<<(v-1));
            setv[o] = v;
            return ;
        }
        
        int M = L+(R-L)/2;
        pushdown(o, L, R);
        
        if(y1 <= M) update(lc, L, M, y1, y2, v);
        if(y2 > M) update(rc, M+1, R, y1, y2, v);
        pushup(o);
    }
    
    int query(int o, int L, int R, int y1, int y2)
    {
        if(y1 <= L && y2 >= R) return color[o];
        
        int M = L+(R-L)/2;
        pushdown(o, L, R);
        
        int ans = 0;
        if(y1 <= M) ans |= query(lc, L, M, y1 ,y2);
        if(y2 > M) ans |= query(rc, M+1, R, y1, y2);
        return ans;
    }
    
    void print(int o, int L, int R)
    {
        printf("%d ", color[o]);
        if(L == R) return ;
        
        int M = L+(R-L)/2;
        print(lc, L, M); print(rc, M+1, R);
    }
};

////////////////

void readint(int &x)
{
    char c = getchar();
    while(!isdigit(c)) c = getchar();
    
    x = 0;
    while(isdigit(c))
    {
        x = x*10+c-'0';
        c = getchar();
    }
}

void writeint(int x)
{
    if(x > 9) writeint(x/10);
    printf("%d", x%10);
}

SegmentTree g;
int n, m, c;

int main()
{
    while(~scanf("%d%d", &n, &m))
    {
        if(n == 0 && m == 0) break;
        g.init(n+10);
        g.build(1, 1, n);
        while(m--)
        {
            char cmd[10]; int x, y, z;
            scanf("%s", cmd);
            if(cmd[0] == 'P')
            {
                readint(x), readint(y), readint(z);
                if(x > y)
                    swap(x, y);
                g.update(1, 1, n, x, y, z);
            }
            else if(cmd[0] == 'Q')
            {
                readint(x), readint(y);
                if(x > y)
                    swap(x, y);
                int tmp = g.query(1, 1, n, x, y);
                
                int ans = 1;
                
                while(tmp)
                {
                    if(tmp & 1 && tmp/2 != 0) writeint(ans), putchar(' ');
                    else if(tmp & 1) writeint(ans), puts("");
                    tmp >>= 1;
                    ans++;
                }
            }
        }
    }
    return 0;
}

HDU 5204：给定一个迷宫，求最长的一条路径，该路径上任意一点都可以通过最多不超过一次的转弯到达。

 

BFS。一直往一个方向走，初始转弯数为-1。元素出队列转弯数+1，每次将转弯数为0或者1的入队，对所有队列中的元素，标记走过坐标的时间。可以证明，经过0次或者1次转弯到达的地方，最小到达的时间即为出队列的时间。枚举所有'.'为起点，取最小到达时间的最大值。

 

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

int n, m;

const int maxn = 110;

char map[maxn][maxn];

int vis[maxn][maxn];

struct node
{
    int x, y;
    int step, k;
};

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

const int dx2_[] = {0, 0, 1, -1};
const int dy2_[] = {1, -1, 0, 0};

int check(int x, int y)
{
    if(x >= 0 && x < n && y >= 0 && y < m && map[x][y] != '#')
        return 1;
    return 0;
}

int bfs(int bx, int by)
{
    memset(vis, 0, sizeof(vis));
    
    queue<node> Q;
    
    node cur, next;
    cur.x = bx, cur.y = by, cur.step = 1, cur.k = -1;
    Q.push(cur);
    vis[bx][by] = 1;
    
    while(!Q.empty())
    {
        cur = Q.front(); Q.pop();
        
        cur.k++;
        int x = cur.x, y = cur.y, k = cur.k;
        if(k == 0)
        {
            for(int i = 0; i < 8; i++)
            {
                next = cur;
                int step = cur.step;
                
                int newx = x+dx[i], newy = y+dy[i];
            
                while(check(newx, newy) && !vis[newx][newy])
                {
                    step++;
                    if(!vis[newx][newy])
                    {
                        vis[newx][newy] = step;
                        next.x = newx, next.y = newy; next.step = step; next.k = k;
                        Q.push(next);
                    }
                    newx += dx[i], newy += dy[i];
                }
            }
        }
        else if(k == 1)
        {
            int x = cur.x, y = cur.y, k = cur.k;
            for(int i = 0; i < 4; i++)
            {
                next = cur;
                int step = cur.step;
                int newx = x+dx2_[i], newy = y+dy2_[i];
                
                while(check(newx, newy) && !vis[newx][newy])
                {
                    step++;
                    if(!vis[newx][newy])
                    {
                        vis[newx][newy] = step;
                    }
                    newx += dx2_[i], newy += dy2_[i];
                }
            }
        }
    }
    
    int ans = 0;
    for(int i = 0; i < n; i++)
    for(int j = 0; j < m; j++) if(vis[i][j])
        ans = max(ans, vis[i][j]);
    return ans;
}

int main()
{
    while(~scanf("%d", &n) && n)
    {
        m = n;
        for(int i = 0; i < n; i++)
            scanf("%s", map[i]);
        
        int ans = 0;
        
        for(int i = 0; i < n; i++)
        for(int j = 0; j < m; j++)
            if(map[i][j] == '.')
                ans = max(ans, bfs(i, j));
        printf("%d\n", ans);
    }
    return 0;
}

HDU 5205：你是孙悟空，给定一个迷宫，起点、终点。迷宫中有M把钥匙，编号分别为1-9，每次只能向相邻的四个点走，每次花费1时间，如果有蛇'S'，花费2时间，杀死之后不再花费时间，只有按照顺序取完所有的钥匙你才有可能救走唐僧，求最小的时间。

 

影响当前节点状态的当前坐标、花费时间、所取得钥匙以及杀死的蛇，所以开个Node节点把上述的状态存下来。因为当前状态只与三种状态有关，坐标、钥匙，会有很多无意义的重复状态。所以标记数组开三维的。由于杀蛇的额外时间会导致BFS步数改变，所以用优先队列。用二进制模拟取得的钥匙，刚开始用二进制存杀死的蛇，WA了一次发现爆int了，于是用vector存下杀死的蛇，最后走到终点并且取得所有钥匙即可。

我们队手速快，三题就出线了smiley。

 

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <queue>
#include <cmath>
#include <set>
using namespace std;

const int maxn = 110;
const int INF = 0x3f3f3f3f;

char map[maxn][maxn];

bool vis[maxn][maxn][1<<10];
bool hash[maxn*maxn];

int n, m, K;
int bx, by, ex, ey;

struct Node
{
	int x, y, step;
	int status;
	
	vector<int> Snake;
	bool operator <(const Node &rhs) const
	{
		return step > rhs.step;
	}
};

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

int check(int x, int y)
{
	if(x >= 0 && x < n && y >= 0 && y < m && map[x][y] != '#')
		return 1;
	return 0;
}

int ok(int x, int status)
{
	for(int i = x-2; i >= 0; i--)
	{
		int tmp = status;
		if((tmp & (1<<i)) == 0) return 0;
	}
	return 1;
}

int idx(int x, int y, Node rhs)
{
	int tmp = x*m+y;
	
	int size = rhs.Snake.size();
	for(int i = 0; i < size; i++)
		if(rhs.Snake[i] == tmp) return 1;
	return 0;
}

int bfs()
{
	priority_queue<Node> Q;
	
	Node cur, next;
	
	cur.x = bx, cur.y = by, cur.status = 0, cur.step = 0;
	Q.push(cur);
	
	vis[bx][by][cur.status] = 1;

	while(!Q.empty())
	{
		cur = Q.top(); Q.pop();
		
		int x = cur.x, y = cur.y, step = cur.step;
		if(cur.x == ex && cur.y == ey && cur.status == ((1<<K)-1))
			return step;
		
		for(int i = 0; i < 4; i++)
		{
			next = cur;
			int newx = x+dx[i], newy = y+dy[i];
			
			int status = cur.status;
			if(!check(newx, newy) || vis[newx][newy][status]) continue ;
			
			if(isdigit(map[newx][newy]))
			{
				int tmp = map[newx][newy]-'0';
				
				if(ok(tmp, status))
				{
					next.x = newx, next.y = newy, next.step++;
					status |= (1<<(tmp-1));
					next.status = status;
					Q.push(next);
					vis[newx][newy][next.status] = 1;
				}
				else
				{
					next.x = newx, next.y = newy, next.step++;
					next.status = status;
					Q.push(next);
					vis[newx][newy][status] = 1;
				}
			}
			else
			{
				next.x = newx, next.y = newy; next.status = status;
				if(map[newx][newy] == 'S')
				{
					if(hash[newx*m+newy] && idx(newx, newy, next))
						next.step++;
					else { next.step += 2; next.Snake.push_back(newx*m+newy); }
				}
				else next.step++;
				Q.push(next);
				vis[newx][newy][status] = 1;
			}
		}
	}
	return -1;
}

void init()
{
	memset(hash, 0, sizeof(hash));
	memset(vis, 0, sizeof(vis));
}

int main()
{
	while(~scanf("%d%d", &n, &K) && (n+K))
	{
		init();
		m = n;
		
		for(int i = 0; i < n; i++)
			scanf("%s", map[i]);
			
		for(int i = 0; i < n; i++)
		for(int j = 0; j < m; j++)
			if(map[i][j] == 'K') bx = i, by = j;
			else if(map[i][j] == 'T') ex = i, ey = j;
			else if(map[i][j] == 'S') hash[i*m+j] = 1;
		
		int ans = bfs();
		
		if(ans != -1)
			printf("%d\n", ans);
		else printf("impossible\n");
	}
	return 0;
}
Small_feed 评论 (97)
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[回复]
CHEATBEATER 说:
2014年10月30日 21:12

每人一题还是三个人一起弄三道题？

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[回复]
Avatar_small
Jack Gao 说:
2014年10月30日 22:01

@CHEATBEATER: 我们是弱队。三个人一起写题，我一个人写了三道题。题目难度梯度大，我们运气好就出线了。

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@Jack Gao:哦！挺厉害的！

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