Codeforces Round #192 (Div. 1) B. Biridian Forest 暴力bfs

B. Biridian Forest

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/329/problem/B

Description

You're a mikemon breeder currently in the middle of your journey to become a mikemon master. Your current obstacle is go through the infamous Biridian Forest.

The forest

The Biridian Forest is a two-dimensional grid consisting of r rows and c columns. Each cell in Biridian Forest may contain a tree, or may be vacant. A vacant cell may be occupied by zero or more mikemon breeders (there may also be breeders other than you in the forest). Mikemon breeders (including you) cannot enter cells with trees. One of the cells is designated as the exit cell.

The initial grid, including your initial position, the exit cell, and the initial positions of all other breeders, will be given to you. Here's an example of such grid (from the first example):

Moves

Breeders (including you) may move in the forest. In a single move, breeders may perform one of the following actions:

• Do nothing.
• Move from the current cell to one of the four adjacent cells (two cells are adjacent if they share a side). Note that breeders cannot enter cells with trees.
• If you are located on the exit cell, you may leave the forest. Only you can perform this move — all other mikemon breeders will never leave the forest by using this type of movement.

After each time you make a single move, each of the other breeders simultaneously make a single move (the choice of which move to make may be different for each of the breeders).

Mikemon battle

If you and t (t > 0) mikemon breeders are located on the same cell, exactly t mikemon battles will ensue that time (since you will be battling each of those t breeders once). After the battle, all of those t breeders will leave the forest to heal their respective mikemons.

Note that the moment you leave the forest, no more mikemon battles can ensue, even if another mikemon breeder move to the exit cell immediately after that. Also note that a battle only happens between you and another breeders — there will be no battle between two other breeders (there may be multiple breeders coexisting in a single cell).

Your goal

You would like to leave the forest. In order to do so, you have to make a sequence of moves, ending with a move of the final type. Before you make any move, however, you post this sequence on your personal virtual idol Blog. Then, you will follow this sequence of moves faithfully.

Goal of other breeders

Because you post the sequence in your Blog, the other breeders will all know your exact sequence of moves even before you make your first move. All of them will move in such way that will guarantee a mikemon battle with you, if possible. The breeders that couldn't battle you will do nothing.

Your task

Print the minimum number of mikemon battles that you must participate in, assuming that you pick the sequence of moves that minimize this number. Note that you are not required to minimize the number of moves you make.

Under two situations the player could score one point.

⋅1. If you touch a buoy before your opponent, you will get one point. For example if your opponent touch the buoy #2 before you after start, he will score one point. So when you touch the buoy #2, you won't get any point. Meanwhile, you cannot touch buoy #3 or any other buoys before touching the buoy #2.

⋅2. Ignoring the buoys and relying on dogfighting to get point. If you and your opponent meet in the same position, you can try to fight with your opponent to score one point. For the proposal of game balance, two players are not allowed to fight before buoy #2 is touched by anybody.

There are three types of players.

Speeder: As a player specializing in high speed movement, he/she tries to avoid dogfighting while attempting to gain points by touching buoys.
Fighter: As a player specializing in dogfighting, he/she always tries to fight with the opponent to score points. Since a fighter is slower than a speeder, it's difficult for him/her to score points by touching buoys when the opponent is a speeder.
All-Rounder: A balanced player between Fighter and Speeder.

There will be a training match between Asuka (All-Rounder) and Shion (Speeder).
Since the match is only a training match, the rules are simplified: the game will end after the buoy #1 is touched by anybody. Shion is a speed lover, and his strategy is very simple: touch buoy #2,#3,#4,#1 along the shortest path.

Asuka is good at dogfighting, so she will always score one point by dogfighting with Shion, and the opponent will be stunned for T seconds after dogfighting. Since Asuka is slower than Shion, she decides to fight with Shion for only one time during the match. It is also assumed that if Asuka and Shion touch the buoy in the same time, the point will be given to Asuka and Asuka could also fight with Shion at the buoy. We assume that in such scenario, the dogfighting must happen after the buoy is touched by Asuka or Shion.

The speed of Asuka is V1 m/s. The speed of Shion is V2 m/s. Is there any possibility for Asuka to win the match (to have higher score)?

Input

The first line consists of two integers: r and c (1 ≤ r, c ≤ 1000), denoting the number of rows and the number of columns in Biridian Forest. The next r rows will each depict a row of the map, where each character represents the content of a single cell:

• 'T': A cell occupied by a tree.
• 'S': An empty cell, and your starting position. There will be exactly one occurence of this in the map.
• 'E': An empty cell, and where the exit is located. There will be exactly one occurence of this in the map.
• A digit (0-9): A cell represented by a digit X means that the cell is empty and is occupied by X breeders (in particular, if X is zero, it means that the cell is not occupied by any breeder).

It is guaranteed that it will be possible for you to go from your starting position to the exit cell through a sequence of moves.

.

Output

A single line denoted the minimum possible number of mikemon battles that you have to participate in if you pick a strategy that minimize this number.

Sample Input

5 7
000E0T3
T0TT0T0
010T0T0
2T0T0T0
0T0S000

Sample Output

3

HINT

 

题意

题目非常长，简单理解下就是给你一个n*m的图，S表示起点，E表示终点，有数字的就是表示有多少个人在那个格子里面

然后问你有多少个人可以比你先到E这个终点

题解：

由终点开始直接爆搜就好了

代码

 

#include<iostream>
#include<stdio.h>
#include<queue>
using namespace std;

int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
string s[1005];
int d[1005][1005];
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++)
        cin>>s[i];
    pair<int,int> now;
    pair<int,int> goal;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            d[i][j]=1005*1005;
            if(s[i][j]=='E')
            {
                now.first = i;
                now.second = j;
                d[i][j]=0;
            }
            if(s[i][j]=='S')
            {
                goal.first = i;
                goal.second = j;
            }
        }
    }
    queue<pair<int,int> > Q;
    Q.push(now);
    while(!Q.empty())
    {
        now = Q.front();
        Q.pop();
        for(int i=0;i<4;i++)
        {
            pair<int,int> next = now;
            next.first += dx[i];
            next.second += dy[i];
            if(next.first<0||next.first>=n)
                continue;
            if(next.second<0||next.second>=m)
                continue;
            if(s[next.first][next.second]=='T')
                continue;
            if(d[next.first][next.second] > d[now.first][now.second] + 1)
            {
                d[next.first][next.second] = d[now.first][now.second] + 1;
                Q.push(next);
            }
        }
    }
    int ans = 0;
    int pp = d[goal.first][goal.second];
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            //cout<<d[i][j]<<" ";
            if(s[i][j]<='9'&&s[i][j]>='0'&&d[i][j]-1<pp)
            {
                ans += s[i][j]-'0';
            }
        }
        //cout<<endl;
    }
    printf("%d\n",ans);
}