UVA - 434 Matty's Blocks

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題意：給你正視和側視圖，求最多多少個，最少多少個

思路：貪心的思想，求最少的時候：因為可以想象著移動，盡量讓兩個視圖的重疊，所以我們統計每個視圖不同高度的個數，然後計算，至於的話，就是每次拿正視圖的高度去匹配側視求最大

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;const int MAXN = 1000;int k;int view[2][MAXN];int main() {int t;scanf("%d", &t);while (t--) {scanf("%d", &k);memset(view, 0, sizeof(view));for (int i = 0; i < 2; i++)for (int j = 0; j < k; j++) {int x;scanf("%d", &x);view[i][x]++;}int Min = 0, Max = 0;for (int i = 1; i < MAXN; i++) Min += i * max(view[0][i], view[1][i]);for (int i = 1; i < MAXN; i++)for (int j = 1; j < MAXN; j++) Max += min(i, j)*view[0][i]*view[1][j];printf("Matty needs at least %d blocks, and can add at most %d extra blocks.\n", Min, Max-Min);}return 0;}