zoukankan      html  css  js  c++  java
  • HDU 3338 Kakuro Extension (最大流)

    Kakuro Extension

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

    Description
    If you solved problem like this, forget it.Because you need to use a completely different algorithm to solve the following one.
    Kakuro puzzle is played on a grid of "black" and "white" cells. Apart from the top row and leftmost column which are entirely black, the grid has some amount of white cells which form "runs" and some amount of black cells. "Run" is a vertical or horizontal maximal one-lined block of adjacent white cells. Each row and column of the puzzle can contain more than one "run". Every white cell belongs to exactly two runs — one horizontal and one vertical run. Each horizontal "run" always has a number in the black half-cell to its immediate left, and each vertical "run" always has a number in the black half-cell immediately above it. These numbers are located in "black" cells and are called "clues".The rules of the puzzle are simple: 

    1.place a single digit from 1 to 9 in each "white" cell
    2.for all runs, the sum of all digits in a "run" must match the clue associated with the "run"

    Given the grid, your task is to find a solution for the puzzle.
                  
            Picture of the first sample input            Picture of the first sample output
     
    Input
    The first line of input contains two integers n and m (2 ≤ n,m ≤ 100) — the number of rows and columns correspondingly. Each of the next n lines contains descriptions of m cells. Each cell description is one of the following 7-character strings: 

    .......— "white" cell;
    XXXXXXX— "black" cell with no clues;
    AAABBB— "black" cell with one or two clues. AAA is either a 3-digit clue for the corresponding vertical run, or XXX if there is no associated vertical run. BBB is either a 3-digit clue for the corresponding horizontal run, or XXX if there is no associated horizontal run.
    The first row and the first column of the grid will never have any white cells. The given grid will have at least one "white" cell.It is guaranteed that the given puzzle has at least one solution.
     
    Output
    Print n lines to the output with m cells in each line. For every "black" cell print '_' (underscore), for every "white" cell print the corresponding digit from the solution. Delimit cells with a single space, so that each row consists of 2m-1 characters.If there are many solutions, you may output any of them.
     
    Sample Input
    6 6
    XXXXXXX XXXXXXX 028XXX 017XXX 028XXX XXXXXXX
    XXXXXXX 02222 ....... ....... ....... 010XXX
    XXX34 ....... ....... ....... ....... .......
    XXX14 ....... ....... 01613 ....... .......
    XXX22 ....... ....... ....... ....... XXXXXXX
    XXXXXXX XXX16 ....... ....... XXXXXXX XXXXXXX
    5 8
    XXXXXXX 001XXX 020XXX 027XXX 021XXX 028XXX 014XXX 024XXX
    XXX35 ....... ....... ....... ....... ....... ....... .......
    XXXXXXX 00734 ....... ....... ....... ....... ....... .......
    XXX43 ....... ....... ....... ....... ....... ....... .......
    XXX30 ....... ....... ....... ....... ....... ....... XXXXXXX
     
    Sample Output
    _ _ _ _ _ _
    _ _ 5 8 9 _
    _ 7 6 9 8 4
    _ 6 8 _ 7 6
    _ 9 2 7 4 _
    _ _ 7 9 _ _
    _ _ _ _ _ _ _ _
    _1 9 9 1 1 8 6
    _ _ 1 7 7 9 1 9
    _ 1 3 9 9 9 3 9
    _ 6 7 2 4 9 2 _

     分析

     神奇的最大流

     可以将左边的有数字的黑色格子看做源点,空白格看做节点,上面有数字的黑色格子看做汇点,跑最大流。

     填入的数字是1~9,可以减一,就是容量为0~8,最后再加回来。

     

    #include <stdio.h>
    #include <string.h>
    #include <iostream>
    #include <algorithm>
    using namespace std;
    
    const int MAXN=20010;
    const int MAXE=200010;
    const int INF=0x3f3f3f3f;
    struct Node
    {
        int to,next,cap;
    }edge[MAXE];
    int tol;
    int head[MAXN];
    int gap[MAXN],dis[MAXN],pre[MAXN],cur[MAXN];
    void init()
    {
        tol=0;
        memset(head,-1,sizeof(head));
    }
    void addedge(int u,int v,int w,int rw=0)
    {
        edge[tol].to=v;edge[tol].cap=w;edge[tol].next=head[u];head[u]=tol++;
        edge[tol].to=u;edge[tol].cap=rw;edge[tol].next=head[v];head[v]=tol++;
    }
    int sap(int start,int end,int nodenum)
    {
        memset(dis,0,sizeof(dis));
        memset(gap,0,sizeof(gap));
        memcpy(cur,head,sizeof(head));
        int u=pre[start]=start,maxflow=0,aug=-1;
        gap[0]=nodenum;
        while(dis[start]<nodenum)
        {
            loop:
            for(int  &i=cur[u];i!=-1;i=edge[i].next)
            {
                int v=edge[i].to;
                if(edge[i].cap&&dis[u]==dis[v]+1)
                {
                    if(aug==-1||aug>edge[i].cap)
                        aug=edge[i].cap;
                    pre[v]=u;
                    u=v;
                    if(v==end)
                    {
                        maxflow+=aug;
                        for(u=pre[u];v!=start;v=u,u=pre[u])
                        {
                            edge[cur[u]].cap-=aug;
                            edge[cur[u]^1].cap+=aug;
                        }
                        aug=-1;
                    }
                    goto loop;
                }
            }
            int mindis=nodenum;
            for(int i=head[u];i!=-1;i=edge[i].next)
            {
                int v=edge[i].to;
                if(edge[i].cap&&mindis>dis[v])
                {
                    cur[u]=i;
                    mindis=dis[v];
                }
            }
            if((--gap[dis[u]])==0)break;
            gap[dis[u]=mindis+1]++;
            u=pre[u];
        }
        return maxflow;
    }
    
    char str[110][110][10];
    int id[110][110];
    int lx[110][110];
    int ly[110][110];
    int num[20010];
    
    int main()
    {
        int n,m;
        while(scanf("%d%d",&n,&m)!=EOF){
            for(int i=0;i<n;i++)
                for(int j=0;j<m;j++)
                    scanf("%s",str[i][j]);
            init();
            int tt=0;
            memset(lx,0,sizeof(lx));
            memset(ly,0,sizeof(ly));
            memset(num,0,sizeof(num));
            for(int i=0;i<n;i++){
                for(int j=0;j<m;j++){
                    if(strcmp(str[i][j],".......")==0){
                        if(j==0||lx[i][j-1]==0) lx[i][j]=++tt;
                        else lx[i][j]=lx[i][j-1];
                        num[lx[i][j]]++;
                    }
                }
            }
            for(int j=0;j<m;j++){
                for(int i=0;i<n;i++){
                    if(strcmp(str[i][j],".......")==0){
                        if(i==0||ly[i-1][j]==0) ly[i][j]=++tt;
                        else ly[i][j]=ly[i-1][j];
                        num[ly[i][j]]++;
                    }
                }
            }
            int st=0,ed=tt+1,nodenum=tt+2;
            for(int i=0;i<n;i++){
                for(int j=0;j<m;j++){
                    if(strcmp(str[i][j],".......")==0){
                       addedge(lx[i][j],ly[i][j],8);
                       id[i][j]=tol-2;
                    }
                }
            }
            for(int i=0;i<n;i++){
                for(int j=0;j<m;j++){
                    if(str[i][j][3]!='\') continue;
                    if(str[i][j][0]!='X'){
                        int tmp=(str[i][j][0]-'0')*100+(str[i][j][1]-'0')*10+(str[i][j][2]-'0');
                        if(ly[i+1][j]!=0) addedge(ly[i+1][j],ed,tmp-num[ly[i+1][j]]);
                    }
                    if(str[i][j][4]!='X'){
                        int tmp=(str[i][j][4]-'0')*100+(str[i][j][5]-'0')*10+(str[i][j][6]-'0');
                        if(lx[i][j+1]!=0) addedge(st,lx[i][j+1],tmp-num[lx[i][j+1]]);
                    }
                }
            }
            sap(st,ed,nodenum);
            for(int i=0;i<n;i++){
                for(int j=0;j<m;j++){
                    if(j>0) printf(" ");
                    if(strcmp(str[i][j],".......")!=0) printf("_");
                    else printf("%c",'0'+9-edge[id[i][j]].cap);
                }
                printf("
    ");
            }
        }
        return 0;
    }
  • 相关阅读:
    server.c:5170:31: error: ‘struct redisServer'
    SpringBoot配置文件笔记:yaml语法,yaml如何写、注入配置文件类(2种方式根据业务场景选择)、多环境切换、配置文件加载优先级、指定位置加载配置文件
    微服务(Microservices)—— Martin Flower
    利用原生子窗体解决悬浮窗口播放的问题及踩坑记录
    uniapp中nvue页面如何使用iconfont字体图标
    项目经验踩坑记录:跨平台业务影响时注意要考虑多个平台
    SQL基础知识笔记:概述(层状/网状/关系模型)、数据类型、操作数据库能力(DDL/DML/DQL)、关系模型(主键、联合主键、外键、外键约束-性能影响、一对一、一对多、多对多、索引、索引效率)、实用SQL语句、事务(四个特性、四种隔离级别)
    Java的Maven基础知识笔记:Maven是什么、maven目录、pom.xml唯一ID、maven解决依赖管理、maven中央仓库与镜像、构建流程(声明周期、阶段、目标)、使用插件、模块管理、mvnw指定版本、如何发布自己的开源库
    Java里的IO基础知识笔记:IO流、字节流/字符流、File对象读取、输入流/输出流(使用过后及时关闭、缓冲区)、Filter模式、ZIP操作、读取classpath资源的意义、序列化/反序列化、Reader/Writer、使用Files工具类及其局限性
    推荐了解一个用于JavaScript的快速SQL数据库
  • 原文地址:https://www.cnblogs.com/wangdongkai/p/5623861.html
Copyright © 2011-2022 走看看