HDU 3392 Pie(DP)

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题意：n男m女，匹配最多对，求最小身高差和。

思路：由于|n - m| <= 100，所以可以用dp[i][j]表示人少的数列的第i个匹配到另一数列的i + j - 1个时的最小身高差和。显然，由于我们要优先匹配多的对，所以对于人少数列的第i个匹配第二队列的人的下标一定是大于等于i的。因此有转移方程: dp[i][j] = max(dp[i-1][j] + fabs(a[i] - b[i+j-1]), dp[i][j-1])。可以用滚动数组优化空间。

#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <string>#include <queue>#include <stack>#include <cmath>#include <set>#include <map>using namespace std;typedef long long LL;#define mem(a, n) memset(a, n, sizeof(a))#define rep(i, n) for(int i = 0; i < (n); i ++)#define REP(i, t, n) for(int i = (t); i < (n); i ++)#define FOR(i, t, n) for(int i = (t); i <= (n); i ++)#define ALL(v) v.begin(), v.end()#define si(a) scanf("%d", &a)#define sii(a, b) scanf("%d%d", &a, &b)#define siii(a, b, c) scanf("%d%d%d", &a, &b, &c)#define pb push_back#define eps 1e-8const int inf = 0x3f3f3f3f, N = 1e4 + 5, MOD = 1e9 + 7;int T, cas = 0;int n, m;double a[N], b[N], dp[2][105];void swap() {if(n > m) {for(int i = 1; i <= n; i ++) swap(a[i], b[i]);swap(n, m);}}int main(){#ifdef LOCAL    freopen("/Users/apple/input.txt", "r", stdin);//freopen("/Users/apple/out.txt", "w", stdout);#endif    while(sii(n, m), n + m) {    mem(a, 0), mem(b, 0);    for(int i = 1; i <= n; i ++) scanf("%lf", &a[i]);    for(int i = 1; i <= m; i ++) scanf("%lf", &b[i]);    sort(a + 1, a + 1 + n);    sort(b + 1, b + 1 + m);    swap();    mem(dp, 0x3f);    int len = m - n + 1;    for(int i = 1; i <= n; i ++) {    for(int j = 1; j <= len; j ++) {    int k = i + j - 1;    dp[i&1][j] = min(dp[(i-1)&1][j] + fabs(a[i] - b[k]), dp[i&1][j-1]);    }    }    printf("%.6f\n", dp[n&1][len]);    }        return 0;}

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