UVALive - 4636 Cubist Artwork

题意：给你正视图，和侧视图，求最好的组成放块个数

思路：当正视图的对应部分跟测视图的对应部分相等的时候，那么就可以当成一个来处理

题目的图有给出这一细节

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 30;

int num1[MAXN],num2[MAXN];
int n,m;

int main(){
    while (scanf("%d%d",&m,&n) != EOF && n+m){
        memset(num1,0,sizeof(num1));
        memset(num2,0,sizeof(num2));
        int x;
        for (int i = 0; i < m; i++){
            scanf("%d",&x);
            num1[x]++;
        }
        for (int i = 0; i < n; i++){
            scanf("%d",&x);
            num2[x]++;
        }
        int ans = 0;
        for (int i = 1; i < 25; i++)
            ans += i*max(num1[i],num2[i]);
        printf("%d\n",ans);
    }
    return 0;
}

UVALive - 4636 Cubist Artwork

原文：http://blog.csdn.net/u011345136/article/details/18451633