hdu 2258 优先队列

Continuous Same Game (1)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 410    Accepted Submission(s): 143

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 5

35552

31154

33222

21134

12314

 

Sample Output

32

 

优先队列，取反，记住！！

做了hdu3090，这道题就很有思路的。

 

题意：题目要求每一次按照贪心的步骤来做，所以只要根据贪心的规则来模拟这个过程就好了。

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

int n,m,MAX;
char tom[20][20];
int a[20][20];
int to[4][2]={{1,0},{0,1},{0,-1},{-1,0}};
bool hash[20][20];
bool use[20][20];
struct node
{
    friend bool operator< (node n1,node n2)
    {
        if(n1.val>n2.val)return false;//no return true;
        else if(n1.val==n2.val)
        {
            if(n1.x<n2.x)return false;
            else if(n1.x==n2.x && n1.y<n2.y) return false;
        }
        return true;
    }
    int x,y,val;
};
struct st
{
    int x,y;
};
priority_queue<node>Q;

int bfs(int x,int y,int num)
{
    int i,cout=1;
    queue<st>S;
    st t,cur;
    t.x=x;
    t.y=y;
    hash[x][y]=true;
    S.push(t);

    while(!S.empty())
    {
        cur=S.front();
        S.pop();
        for(i=0;i<4;i++)
        {
            t=cur;
            t.x=t.x+to[i][0];
            t.y=t.y+to[i][1];
            if(t.x>=0&&t.x<n &&t.y>=0&&t.y<m && !hash[t.x][t.y] && a[t.x][t.y]==num){
                hash[t.x][t.y]=true;
                cout++;
                S.push(t);
            }
        }
    }
    return cout;
}
void change(node &t)
{
    int i,j,k;
    int qq[20][20];
    memset(qq,0,sizeof(qq));
    memset(hash,false,sizeof(hash));
    k = bfs(t.x,t.y,a[t.x][t.y]);
    for(i=0;i<n;i++)
        for(j=0;j<m;j++)
            if(hash[i][j]) a[i][j]=0;
    for(i=0;i<m;i++){
        for(j=n-1,k=n-1;j>=0;j--)
            if(a[j][i]) qq[k--][i] = a[j][i];
    }
    memset(a,0,sizeof(a));
    for(i=0,k=0;i<m;i++){
        for(j=0;j<n;j++) if(qq[j][i]!=0)break;
        if(j==n)continue;
        for(j=n-1;j>=0;j--)
            a[j][k]=qq[j][i];
        k++;
    }
}
void dfs(int now)
{
    int i,j,val;
    bool flag=false;
    node t;
    while(!Q.empty()){
        Q.pop();
    }
    if(now>MAX) MAX = now;
    memset(hash,false,sizeof(hash));
    for(i=0;i<n;i++){
        for(j=0;j<m;j++){
            if(!hash[i][j] && a[i][j]!=0)
            {
                val = bfs(i,j,a[i][j]);
                if(val == 1) continue;
                t.x=i;
                t.y=j;
                t.val=val*(val-1);
                flag=true;
                Q.push(t);
            }
        }
    }
    if(flag==false) return;
    t=Q.top();
    change(t);
    dfs(now+t.val);
}
int main()
{
    int i,j;
    while(scanf("%d%d",&n,&m)>0)
    {
        for(i=0;i<n;i++)
            scanf("%s",tom[i]);
        for(i=0;i<n;i++)
            for(j=0;j<m;j++)
                a[i][j]=tom[i][j]-'0';
        MAX=-1;
        dfs(0);
        printf("%d
",MAX);
    }
    return 0;
}