【bzoj1611】【Usaco2008 Feb】Meteor Shower流星雨（bfs）题解&代码

最新推荐文章于 2020-04-12 11:55:38 发布

deritt

最新推荐文章于 2020-04-12 11:55:38 发布

阅读量690

点赞数

分类专栏： bzoj 文章标签： usaco bzoj

本文链接：https://blog.youkuaiyun.com/DERITt/article/details/51426173

版权

bzoj 专栏收录该内容

18 篇文章

订阅专栏

This blog post provides a detailed explanation and code for solving the Usaco2008 February contest problem, Meteor Shower (bzoj1611), using Breadth First Search (BFS) algorithm. The author shares insights into their approach and presents the implementation to help readers understand the problem-solving process." 135353090,7337247,从知识获取到创造：深度学习时代的人工智能探索,"['深度学习', '机器学习', '神经网络', '自然语言处理', '知识图谱']

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目链接：
http://www.lydsy.com/JudgeOnline/problem.php?id=1611
题解：
直接bfs。。。
代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#define m (300)
using namespace std;
struct node{
    int x;int y;
};
queue<node>q;
node jia(int xx,int yy)
{
    node tmp;
    tmp.x=xx;
    tmp.y=yy;
    return tmp;
}
int n,vis[305][305],dis[305][305];
int d[2][4]={{0,0,-1,1},{1,-1,0,0}};
int main()
{
    for (int i=0;i<=301;i++)
    for (int j=0;j<=301;j++)
    {
        vis[i][j]=999999;
        dis[i][j]=-1;
    }
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
    {
        int x,y,t;
        scanf("%d%d%d",&x,&y,&t);
        vis[x][y]=min(vis[x][y],t);
        for (int j=0;j<4;j++)
        {
            int nex=x+d[0][j];
            int ney=y+d[1][j];
            if (nex>=0&&ney>=0)
            vis[nex][ney]=min(vis[nex][ney],t);
        }
    }
    q.push(jia(0,0));
    dis[0][0]=0;
    if (vis[0][0]==999999)
    {
        printf("0\n");
        return 0;
    }
    if (vis[0][0]<=0)
    {
        printf("-1\n");
        return 0;
    }
    while(!q.empty())
    {
        node now=q.front();q.pop();
        for (int i=0;i<4;i++)
        {
            int nex=now.x+d[0][i];
            int ney=now.y+d[1][i];
            if (nex>=0&&nex<=301&&ney>=0&&ney<=301)
            if (dis[now.x][now.y]+1<vis[nex][ney]&&dis[nex][ney]==-1)
            {
                dis[nex][ney]=dis[now.x][now.y]+1;
                if (vis[nex][ney]==999999)
                {
                    printf("%d\n",dis[nex][ney]);
                    return 0;
                }
                q.push(jia(nex,ney));
            }
        }
    }
    printf("-1\n");
}