• HDU 4451 Dressing


    Problem Description
    Wangpeng has N clothes, M pants and K shoes so theoretically he can have N×M×K different combinations of dressing.
    One day he wears his pants Nike, shoes Adiwang to go to school happily. When he opens the door, his mom asks him to come back and switch the dressing. Mom thinks that pants-shoes pair is disharmonious because Adiwang is much better than Nike. After being asked to switch again and again Wangpeng figure out all the pairs mom thinks disharmonious. They can be only clothes-pants pairs or pants-shoes pairs.
    Please calculate the number of different combinations of dressing under mom’s restriction.
     
    Input
    There are multiple test cases.
    For each case, the first line contains 3 integers N,M,K(1≤N,M,K≤1000) indicating the number of clothes, pants and shoes.
    Second line contains only one integer P(0≤P≤2000000) indicating the number of pairs which mom thinks disharmonious.
    Next P lines each line will be one of the two forms“clothes x pants y” or “pants y shoes z”.
    The first form indicates pair of x-th clothes and y-th pants is disharmonious(1≤x≤N,1 ≤y≤M), and second form indicates pair of y-th pants and z-th shoes is disharmonious(1≤y≤M,1≤z≤K).
    Input ends with “0 0 0”.
    It is guaranteed that all the pairs are different.
     
    Output
    For each case, output the answer in one line.
    Sample Input
    2 2 2
    0
    2 2 2
    1
    clothes 1 pants 1
    2 2 2
    2 clothes 1 pants 1
    pants 1 shoes 1
    0 0 0
    Sample Output
    8
    6
    5
     
    Source

    对于这题,看似跟容斥定理一样,不过P的值太大果断放弃。

    后来想想只有三组结点,结点的个数为1000,不大果断使用使用map记录一个结点到另外一个结点的边。并且中间结点记录出度个数。

    然后删边,最后使用加法原理直接加。

    #include <stdio.h>
    #include <string.h>
    #define MAXN 1001
    
    int N,M,K;
    int map[MAXN][MAXN];
    int out[MAXN];
    int main(int argc, char *argv[])
    {
        char ch1[10],ch2[10];
        int x,y;
        while( scanf("%d %d %d",&N,&M,&K)!=EOF ){
            if(N==0 &&M==0 && K==0)break;
            memset(map,1,sizeof(map));
            for(int i=1; i<=M; i++){
                out[i]=K;
            }
            int P;
            scanf("%d",&P);
            while(P--){
                scanf("%s %d %s %d",ch1,&x,ch2,&y);
                if(ch1[0]=='c' && ch2[0]=='p'){
                    map[x][y]=0;
                }
                if(ch1[0]=='p' && ch2[0]=='s'){
                    if(out[x]>0)
                        out[x]--;
                }
            }
            __int64 sum=0;
            for(int i=1; i<=N; i++){
                for(int j=1; j<=M; j++){
                    if(map[i][j]){
                        sum+=out[j];
                    }
                }
            }
            printf("%I64d
    ",sum);
        }    
        return 0;
    }
  • 相关阅读:
    sonarqube代码质量检测工具安装(docker安装篇)
    jenkins-job构建完成后通知企业微信
    jenkins-构建job成功后自动打tag到git仓库
    初识Python-变量
    初识Python-文件类型(入坑篇)
    kubernetes-部署LNMP环境运行Discuz
    harbor部署常见的错误
    kubernetes-部署harbor
    Google Hacking语法
    信息搜集总结
  • 原文地址:https://www.cnblogs.com/chenjianxiang/p/3552412.html
走看看 - 开发者的网上家园