uva 434 - Matty's Blocks(贪心)
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题目链接:uva 434 - Matty's Blocks
题目大意:给出前视图和右视图,计算出最少需要几个正方体以及至多可再增加几个正方体。
解题思路:和昨天做得一题uva 1445一样的,只是增加了要计算说最多可以放几个正方体,贪心,尽量让大的当着小的。
#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;const int N = 10;int n, f[N], r[N];void init() {memset(f, 0, sizeof(f));memset(r, 0, sizeof(r));scanf("%d", &n);int a;for (int i = 0; i < n; i++) {scanf("%d", &a); f[a]++;}for (int i = 0; i < n; i++) {scanf("%d", &a); r[a]++;}}void solve() {int Min = 0, Max = 0, left = n, right = n;for (int i = 0; i < N; i++) {Min += max(f[i], r[i]) * i;Max += left * right;left -= f[i];right -= r[i];}printf("Matty needs at least %d blocks, and can add at most %d extra blocks.\n", Min, Max - Min - n * n);}int main () {int cas;scanf("%d", &cas);while (cas--) {init();solve();}return 0;}
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