hdoj 2258 Continuous Same Game (1) (DFS+模拟)

没什么意思，，根据题意模拟出过程就好。。

Continuous Same Game (1)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 412    Accepted Submission(s): 145

Problem Description

Continuous Same Game is a simple game played on a grid of colored blocks. Groups of two or more connected (orthogonally, not diagonally) blocks that are the same color may be removed from the board. When a group of blocks is removed, the blocks above those removed ones fall down into the empty space. When an entire column of blocks is removed, all the columns to the right of that column shift to the left to fill the empty columns. Points are scored whenever a group of blocks is removed. The number of points per block increases as the group becomes bigger. When N blocks are removed, N*(N-1) points are scored.

LL was interested in this game at one time, but he found it is so difficult to find the optimal scheme. So he always play the game with a greedy strategy: choose the largest group to remove and if there are more than one largest group with equal number of blocks, choose the one which contains the most preceding block ( (x1,y1) is in front of (x2,y2) if and only if (x1<x2 || x1==x2 && y1<y2) ). Now, he want to know how many points he will get. Can you help him?

 

Input

Each test case begins with two integers n,m ( 5<=n,m<=20 ), which is the size of the board. Then n lines follow, each contains m characters, indicating the color of the block. There are 5 colors, and each with equal probability.

 

Output

For each test case, output a single line containing the total point he will get with the greedy strategy.

 

Sample Input

5 53555231154332222113412314

 

Sample Output

32
Hint
35552    00552    00002    00002    00000    0000031154    05154    05104    00004    00002    0000033222    01222    01222    00122    00104    0010021134    21134    21134    25234    25234    2523012314    12314    12314    12314    12314    12312The total point is 12+6+6+2+6=32.
 

 

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int N = 25;char mp[N][N];int vis[N][N];int dir[4][2] = {1, 0, 0, 1, -1, 0, 0, -1};int n, m;int cnt;void dfs1(int x, int y, char c){    vis[x][y] = 1;    cnt++;    for(int i = 0; i < 4; i++){        int xx = x + dir[i][0];        int yy = y + dir[i][1];        if(mp[xx][yy] == c && !vis[xx][yy]) dfs1(xx, yy, c);    }}void dfs2(int x, int y, char c){    mp[x][y] = 0;    for(int i = 0; i < 4; i++){        int xx = x + dir[i][0];        int yy = y + dir[i][1];        if(mp[xx][yy] == c) dfs2(xx, yy, c);    }}int main(){    while(scanf("%d%d", &n, &m) != EOF){        memset(mp, '\0', sizeof(mp));        for(int i = 1; i <= n; i++){            scanf("%s", mp[i] + 1);            for(int j = 1; j <= m; j++){                if(mp[i][j] == '0') mp[i][j] = 0;            }        }        int ans = 0;        while(true){            memset(vis, 0, sizeof(vis));            int Max = -1, mx, my;            for(int i = 1; i <= n; i++)            for(int j = 1; j <= m; j++){                if(mp[i][j] && !vis[i][j]){                    cnt = 0;                    dfs1(i, j, mp[i][j]);                 //   cout << "cnt = " << cnt << endl;                    if(cnt > Max){                        mx = i; my = j;                        Max = cnt;                    }                }            }            if(Max < 2) break;          //  cout << "Max = " << Max << endl;            ans += Max * (Max - 1);            dfs2(mx, my, mp[mx][my]);            for(int j = 1; j <= m; j++){                int k, flag = 0;                if(mp[n][j]) flag = 1;                for(int i = n - 1; i > 0; i--){                    if(mp[i][j]){                        flag = 1; k = i;                        while(k < n && mp[k + 1][j] == 0){                            mp[k + 1][j] = mp[k][j];                            mp[k][j] = 0;                            k++;                        }                    }                }                if(flag) continue;                for(int i = 1; i <= n; i++)                    for(k = j; k > 0; k--)                        mp[i][k] = mp[i][k - 1];            }        }        printf("%d\n", ans);    }    return 0;}