标签：his diff break common star ext false ber main
《Journey to the West》(also 《Monkey》) is one of the Four Great Classical Novels of Chinese literature. It was written by Wu Cheng‘en during the Ming Dynasty. In this novel, Monkey King Sun Wukong, pig Zhu Bajie and Sha Wujing, escorted Tang Monk to India to get sacred Buddhism texts.
During the journey, Tang Monk was often captured by demons. Most of demons wanted to eat Tang Monk to achieve immortality, but some female demons just wanted to marry him because he was handsome. So, fighting demons and saving Monk Tang is the major job for Sun Wukong to do.
Once, Tang Monk was captured by the demon White Bones. White Bones lived in a palace and she cuffed Tang Monk in a room. Sun Wukong managed to get into the palace, and he wanted to reach Tang Monk and rescue him.
The palace can be described as a matrix of characters. Different characters stand for different rooms as below:
‘S‘ : The original position of Sun Wukong
‘T‘ : The location of Tang Monk
‘.‘ : An empty room
‘#‘ : A deadly gas room.
‘B‘ : A room with unlimited number of oxygen bottles. Every time Sun Wukong entered a ‘B‘ room from other rooms, he would get an oxygen bottle. But staying there would not get Sun Wukong more oxygen bottles. Sun Wukong could carry at most 5 oxygen bottles at the same time.
‘P‘ : A room with unlimited number of speed-up pills. Every time Sun Wukong entered a ‘P‘ room from other rooms, he would get a speed-up pill. But staying there would not get Sun Wukong more speed-up pills. Sun Wukong could bring unlimited number of speed-up pills with him.
Sun Wukong could move in the palace. For each move, Sun Wukong might go to the adjacent rooms in 4 directions(north, west,south and east). But Sun Wukong couldn‘t get into a ‘#‘ room(deadly gas room) without an oxygen bottle. Entering a ‘#‘ room each time would cost Sun Wukong one oxygen bottle.
Each move took Sun Wukong one minute. But if Sun Wukong ate a speed-up pill, he could make next move without spending any time. In other words, each speed-up pill could save Sun Wukong one minute. And if Sun Wukong went into a ‘#‘ room, he had to stay there for one extra minute to recover his health.
Since Sun Wukong was an impatient monkey, he wanted to save Tang Monk as soon as possible. Please figure out the minimum time Sun Wukong needed to reach Tang Monk.
InputThere are no more than 25 test cases.
For each case, the first line includes two integers N and M(0 < N,M ≤ 100), meaning that the palace is a N × M matrix.
Then the N×M matrix follows.
The input ends with N = 0 and M = 0.
OutputFor each test case, print the minimum time (in minute) Sun Wukong needed to save Tang Monk. If it‘s impossible for Sun Wukong to complete the mission, print -1
Sample Input
2 2 S# #T 2 5 SB### ##P#T 4 7 SP..... P#..... ......# B...##T 0 0
Sample Output
-1 8 11
题意：
给你一个二维字符矩阵，S字符代表孙悟空的位置，T字符代表唐生的位置。
现在是孙悟空要用最小的距离去到唐生的位置，
路途中有以下几种位置
‘#‘ 是毒区，要有一一个氧气才可以进去，并且需要额外消耗一个时间来恢复身体。
‘B‘ 是氧气区，进来会拿到一个氧气，每一个氧气区可以无限拿，但是孙悟空身上最多可以拿5个氧气管。
’P‘ 是能量区，进来会获得一个能力包，一个能量包可以让你行动一次不消耗时间。
’.‘ 空区域，任意行走。
思路：
开一个思维的dis数组来记录以下状态
dis[x][y][o][p] 表示 在x，y位置，有o个氧气，p个能力包时的最小用时。
然后用BFS正常搜即可，细节处理下即可。
提示：
每一个能量包能用就用的话可以让代码写起来更简单。
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <map> #include <set> #include <vector> #include <iomanip> #define ALL(x) (x).begin(), (x).end() #define rt return #define dll(x) scanf("%I64d",&x) #define xll(x) printf("%I64d\n",x) #define sz(a) int(a.size()) #define all(a) a.begin(), a.end() #define rep(i,x,n) for(int i=x;i<n;i++) #define repd(i,x,n) for(int i=x;i<=n;i++) #define pii pair<int,int> #define pll pair<long long ,long long> #define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define MS0(X) memset((X), 0, sizeof((X))) #define MSC0(X) memset((X), ‘\0‘, sizeof((X))) #define pb push_back #define mp make_pair #define fi first #define se second #define eps 1e-6 #define gg(x) getInt(&x) #define db(x) cout<<"== [ "<<x<<" ] =="<<endl; using namespace std; typedef long long ll; ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;} ll lcm(ll a, ll b) {return a / gcd(a, b) * b;} ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;} inline void getInt(int* p); const int maxn = 1000010; const int inf = 0x3f3f3f3f; /*** TEMPLATE CODE * * STARTS HERE ***/ int n, m; char s[200][234]; struct node { int x, y, step, o, p; node() {} node(int xx, int yy, int oo, int pp, int ss) { x = xx; y = yy; o = oo; p = pp; step = ss; } }; int xx[] = { -1, 1, 0, 0}; int yy[] = {0, 0, -1, 1}; int dis[104][123][7][5]; int main() { //freopen("D:\\common_text\\code_stream\\in.txt","r",stdin); //freopen("D:\\common_text\\code_stream\\out.txt","w",stdout); while (~scanf("%d %d", &n, &m)) { if (n + m == 0) { break; } repd(i, 1, n) { scanf("%s", s[i] + 1); } int sx, sy; repd(i, 1, n) { repd(j, 1, m) { if (s[i][j] == ‘S‘) { sx = i; sy = j; break; } } } queue<node> q; q.push(node(sx, sy, 0, 0, 0)); memset(dis, inf, sizeof(dis)); // cout<<dis[0][2][2][1]<<endl; int ans = inf; while (!q.empty()) { node temp = q.front(); q.pop(); repd(i, 0, 3) { int tx = temp.x + xx[i]; int ty = temp.y + yy[i]; if (tx >= 1 && tx <= n && ty >= 1 && ty <= m) { if (s[tx][ty] == ‘T‘) { if (temp.p > 0) { dis[tx][ty][temp.o][temp.p] = min(dis[tx][ty][temp.o][temp.p], temp.step); ans = min(dis[tx][ty][temp.o][temp.p], ans); } else { dis[tx][ty][temp.o][temp.p] = min(dis[tx][ty][temp.o][temp.p], temp.step + 1); ans = min(dis[tx][ty][temp.o][temp.p], ans); } } else if (s[tx][ty] == ‘B‘) { // get oxy if (temp.o >= 5) { if (temp.p > 0) { if (dis[tx][ty][temp.o][temp.p] > temp.step) { dis[tx][ty][temp.o][temp.p] = temp.step; q.push(node(tx, ty, temp.o, temp.p - 1, temp.step)); } } else { if (dis[tx][ty][temp.o][temp.p] > temp.step + 1) { dis[tx][ty][temp.o][temp.p] = temp.step + 1; q.push(node(tx, ty, temp.o, temp.p, temp.step + 1)); } } } else { if (temp.p > 0) { if (dis[tx][ty][temp.o][temp.p] > temp.step) { dis[tx][ty][temp.o][temp.p] = temp.step; q.push(node(tx, ty, temp.o + 1, temp.p - 1, temp.step)); } } else { if (dis[tx][ty][temp.o][temp.p] > temp.step + 1) { dis[tx][ty][temp.o][temp.p] = temp.step + 1; q.push(node(tx, ty, temp.o + 1, temp.p, temp.step + 1)); } } } } else if (s[tx][ty] == ‘P‘) { // get p if (temp.p > 0) { if (dis[tx][ty][temp.o][temp.p] > temp.step) { dis[tx][ty][temp.o][temp.p] = temp.step; q.push(node(tx, ty, temp.o, temp.p, temp.step)); } } else { if (dis[tx][ty][temp.o][temp.p] > temp.step + 1) { dis[tx][ty][temp.o][temp.p] = temp.step + 1; q.push(node(tx, ty, temp.o, temp.p + 1, temp.step + 1)); } } } else if (s[tx][ty] == ‘#‘) { // du qu if (temp.o > 0) { if (temp.p > 0) { if (dis[tx][ty][temp.o][temp.p] > temp.step + 1) { dis[tx][ty][temp.o][temp.p] = temp.step + 1; q.push(node(tx, ty, temp.o - 1, temp.p - 1, temp.step + 1)); } } else { if (dis[tx][ty][temp.o][temp.p] > temp.step + 2) { dis[tx][ty][temp.o][temp.p] = temp.step + 2; q.push(node(tx, ty, temp.o - 1, temp.p, temp.step + 2)); } } } } else { // nomal if (temp.p > 0) { if (dis[tx][ty][temp.o][temp.p] > temp.step) { dis[tx][ty][temp.o][temp.p] = temp.step; q.push(node(tx, ty, temp.o, temp.p - 1, temp.step)); } } else { if (dis[tx][ty][temp.o][temp.p] > temp.step + 1) { dis[tx][ty][temp.o][temp.p] = temp.step + 1; q.push(node(tx, ty, temp.o, temp.p, temp.step + 1)); } } } } } } if (ans == inf) { printf("-1\n"); } else { printf("%d\n", ans ); } } return 0; } inline void getInt(int* p) { char ch; do { ch = getchar(); } while (ch == ‘ ‘ || ch == ‘\n‘); if (ch == ‘-‘) { *p = -(getchar() - ‘0‘); while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) { *p = *p * 10 - ch + ‘0‘; } } else { *p = ch - ‘0‘; while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) { *p = *p * 10 + ch - ‘0‘; } } }
Saving Tang Monk II HihoCoder - 1828 2018北京赛站网络赛A题
标签：his diff break common star ext false ber main
原文地址：https://www.cnblogs.com/qieqiemin/p/10807177.html
