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; //
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //
ll mod = ;
const double esp = 1e-;
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=;int f=;char ch=getchar();
while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
return x=f?-x:x;
} /*-----------------------showtime----------------------*/
const int maxn = ;
int n,m;
int sx,sy,ex,ey;
char mp[maxn][maxn];
int dp[maxn][maxn][],vis[maxn][maxn];
int nx[][] = {
{,},{,},{-,},{,-}
};
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=; i<n; i++){
scanf("%s", mp[i]);
for(int j=; 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,));
dp[sx][sy][]=;
while(!que.empty()){
state t = que.front();
que.pop();
for(int i=; i<; i++){
int px = t.x + nx[i][];
int py = t.y + nx[i][];
if(px<||py<||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]+){
dp[px][py][t.cnt] = dp[t.x][t.y][t.cnt]+;
que.push(state(px,py,t.cnt));
}
}
else if(mp[px][py] == '#'){
if(t.cnt&&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-));
}
}
else if(mp[px][py] == 'B'){
if(t.cnt< && 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+));
}
else if(t.cnt == && 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));
}
}
else if(mp[px][py] == 'T'){
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]+;
}
}
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=; i<=; i++){
ans = min(ans, dp[ex][ey][i]);
}
if(ans < inf)printf("%d\n", ans);
else puts("-1");
}
return ;
}
hihocoder 1828