北京2018网络赛 hihocoder#1828 : Saving Tang Monk II （BFS + DP +多开一维）

hihocoder 1828 ：https://hihocoder.com/problemset/problem/1828

学习参考：https://www.cnblogs.com/tobyw/p/9691431.html

题意：

　　　　给定一个图中，让你回答从S点跑到T点的最短时间。“.”点是可以直接走上去，耗时+1，“P”加速点，就是不耗时就可以走上去，“#”毒气点，必须要有氧气瓶才能进入，且耗时+2，“B”是氧气瓶补给点，每次进去可以得到一个氧气瓶，但是你最多可以携带5个氧气瓶，耗时+1。

思路：

　　　　我觉得这道题看到5个氧气瓶的限制是个关键。可以给每个点开5个空间（准确的说是6个，还有0的情况），这样就直接bfs转移就可以了。自己练习的时候一直没想到再多开一维。现在想想，感觉能看见和想到高维度的人都好神奇。

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
// #pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3)

#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
#define max3(a,b,c) max(max(a,b), c); 
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;       
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
ll mod = 10007;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;


template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}


/*-----------------------showtime----------------------*/
            const int maxn = 209;
            int n,m;
            int sx,sy,ex,ey;
            char mp[maxn][maxn];
            int dp[maxn][maxn][10],vis[maxn][maxn];
            int nx[4][4] = {
                {1,0},{0,1},{-1,0},{0,-1}
            };
            struct state
            {
                int x,y,cnt;

                state(int x,int y,int s):x(x),y(y),cnt(s){}
            };
int main(){
            while(~scanf("%d%d", &n, &m) && n+m){
                memset(dp,inf,sizeof(dp));
                for(int i=0; i<n; i++){
                    scanf("%s", mp[i]);
                    for(int j=0; j<m; j++){
                        if(mp[i][j] == 'S')sx=i,sy=j;
                        if(mp[i][j] == 'T')ex=i,ey=j;
                    }
                }
                queue<state>que;
                que.push(state(sx,sy,0));
                dp[sx][sy][0]=0;
                while(!que.empty()){
                    state t = que.front();
                    que.pop();
                    for(int i=0; i<4; i++){
                        int px = t.x + nx[i][0];
                        int py = t.y + nx[i][1];
                        if(px<0||py<0||px>=n||py>=m)continue;
                        // vis[px][py] = 1;
                        if(mp[px][py]=='.' || mp[px][py] == 'S'){
                                if(dp[px][py][t.cnt] > dp[t.x][t.y][t.cnt]+1){
                                    dp[px][py][t.cnt] =  dp[t.x][t.y][t.cnt]+1;
                                    que.push(state(px,py,t.cnt));
                                }
                        }
                        else if(mp[px][py] == '#'){
                            if(t.cnt&&dp[px][py][t.cnt-1] > dp[t.x][t.y][t.cnt]+2){
                                    dp[px][py][t.cnt-1] =  dp[t.x][t.y][t.cnt]+2;
                                    que.push(state(px,py,t.cnt-1));
                            }
                        }
                        else if(mp[px][py] == 'B'){
                            if(t.cnt<5 && dp[px][py][t.cnt+1] > dp[t.x][t.y][t.cnt]+1){
                                    dp[px][py][t.cnt+1] =  dp[t.x][t.y][t.cnt]+1;
                                    que.push(state(px,py,t.cnt+1));
                            }
                            else if(t.cnt == 5 && dp[px][py][t.cnt] > dp[t.x][t.y][t.cnt]+1){
                                    dp[px][py][t.cnt] =  dp[t.x][t.y][t.cnt]+1;
                                    que.push(state(px,py,t.cnt));
                            }
                        }
                        else if(mp[px][py] == 'T'){
                            if(dp[px][py][t.cnt] > dp[t.x][t.y][t.cnt]+1){
                                dp[px][py][t.cnt] = dp[t.x][t.y][t.cnt]+1;
                            }
                        }
                        else if(mp[px][py] == 'P'){
                            if(dp[px][py][t.cnt] > dp[t.x][t.y][t.cnt]){
                                dp[px][py][t.cnt] = dp[t.x][t.y][t.cnt];
                                que.push(state(px,py,t.cnt));
                            }
                        }
                    }
                }
                int ans = inf;
                for(int i=0; i<=5; i++){
                    ans = min(ans, dp[ex][ey][i]);
                }
                if(ans < inf)printf("%d\n", ans);
                else puts("-1");
            }
            return 0;
}