洛谷 P2895 [USACO08FEB]流星雨Meteor Shower

P2895 [USACO08FEB]流星雨Meteor Shower

题目描述

Bessie hears that an extraordinary meteor shower is coming; reports say that these meteors will crash into earth and destroy anything they hit. Anxious for her safety, she vows to find her way to a safe location (one that is never destroyed by a meteor) . She is currently grazing at the origin in the coordinate plane and wants to move to a new, safer location while avoiding being destroyed by meteors along her way.

The reports say that M meteors (1 ≤ M ≤ 50,000) will strike, with meteor i will striking point (Xi, Yi) (0 ≤ Xi ≤ 300; 0 ≤ Yi ≤ 300) at time Ti (0 ≤ Ti  ≤ 1,000). Each meteor destroys the point that it strikes and also the four rectilinearly adjacent lattice points.

Bessie leaves the origin at time 0 and can travel in the first quadrant and parallel to the axes at the rate of one distance unit per second to any of the (often 4) adjacent rectilinear points that are not yet destroyed by a meteor. She cannot be located on a point at any time greater than or equal to the time it is destroyed).

Determine the minimum time it takes Bessie to get to a safe place.

牛去看流星雨，不料流星掉下来会砸毁上下左右中五个点。每个流星掉下的位置和时间都不同，求牛能否活命，如果能活命，最短的逃跑时间是多少？

输入输出格式

输入格式：

 

• Line 1: A single integer: M

• Lines 2..M+1: Line i+1 contains three space-separated integers: Xi, Yi, and Ti

 

输出格式：

 

• Line 1: The minimum time it takes Bessie to get to a safe place or -1 if it is impossible.

 

输入输出样例

输入样例#1：

4
0 0 2
2 1 2
1 1 2
0 3 5

输出样例#1：

5
思路：爆搜。
错因：本题应该用宽搜，用了深搜。
吐槽：大佬竟然是这样判断宽搜深搜的：深搜TLE就是宽搜，宽搜MLE/RE就是深搜╮(╯▽╰)╭。
63分的深搜：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int dx[4]={0,0,-1,1};
int dy[4]={-1,1,0,0};
int ans=0x7f7f7f7f,m,t,map[310][310],mm[310][310];
void dfs(int x,int y,int step){
    if(map[x][y]>2000000000){
        ans=min(ans,step);
        return ;
    }
    for(int i=0;i<4;i++){
        int cx=x+dx[i];
        int cy=y+dy[i];
        if(map[cx][cy]>step&&cx>=0&&cy>=0){
            map[cx][cy]-=1;
            dfs(cx,cy,step+1);
            map[cx][cy]+=1;
        }
    }
}
int main(){
    scanf("%d",&m);
    memset(map,0x7f,sizeof(map));
    for(int i=1;i<=m;i++){
        int x,y,z;
        scanf("%d%d%d",&x,&y,&z);
        map[x][y]=min(map[x][y],z);
        map[x+1][y]=min(map[x+1][y],z);
        map[x][y+1]=min(map[x][y+1],z);
        if(x>0)    map[x-1][y]=min(map[x-1][y],z);
        if(y>0)    map[x][y-1]=min(map[x][y-1],z);
    }
    dfs(0,0,1);
    if(ans!=2139062143)    cout<<ans-1;
    else cout<<"-1";
}

AC的宽搜：

#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define oo 1e9
using namespace std;
struct nond{
    int x,y,z;
};
queue<nond>que;
bool b[310][310];
int n,map[310][310];
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
int main(){
    memset(map,0x7f,sizeof(map));
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        int x,y,z;
        scanf("%d%d%d",&x,&y,&z);
        map[x][y]=min(map[x][y],z);
        map[x+1][y]=min(map[x+1][y],z);
        map[x][y+1]=min(map[x][y+1],z);
        if(x>0)    map[x-1][y]=min(map[x-1][y],z);
        if(y>0)    map[x][y-1]=min(map[x][y-1],z);
    }
    nond now,pre;
    b[0][0]=1;
    now.x=now.y=now.z=0;
    que.push(now);
    while(!que.empty()){
        now=que.front();
        que.pop();
        if(map[now.x][now.y]==2139062143){
            printf("%d",now.z);
            return 0;
        }
        for(int i=0;i<4;i++){
            int cx=now.x+dx[i];
            int cy=now.y+dy[i];
            if(cx>=0&&cy>=0&&!b[cx][cy]&&map[cx][cy]>now.z+1){
                b[cx][cy]=1;
                pre.x=cx;pre.y=cy;pre.z=now.z+1;
                que.push(pre);
            }
        }
    }
    cout<<"-1";
    return 0;
}