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#include #include 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; }