学步园

0-1背包模型，每个人的决策为装 0-j 个storage，j为当前剩余storage的数量

先求出最大的L，再用这个L重新做一次dp，找出最小的Y

Run Time: 0.016s

#define UVa  "9-9.10163.cpp"		//Storage Keepers
char fileIn[30] = UVa, fileOut[30] = UVa;

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<iostream>

using namespace std;

//Global Variables. Reset upon Each Case!
const int maxm = 30 + 5, maxn = 100 + 5, INF = 1<<30;
int n, m, p[maxm];
int d[maxn][maxn];
int d2[maxn][maxn];
int maxL;
/////

int main() {
    while(scanf("%d%d", &n, &m) && n) {
        for(int i = 0; i < m; i ++)
            scanf("%d", &p[i]);
	    
        for(int i = m-1; i >= 0; i --) {
            for(int j = 0; j <= n; j ++) {
                int& u = d[i][j];
                if(j == 0) {
                    u = INF;
                }
                else {
                    if(i == m-1) {
                        u = p[i]/j;
                    }
                    else {
                        u = d[i+1][j];
                        for(int k = 1; k <= j; k ++) {
                            u = max(u, min(p[i]/k, d[i+1][j-k]));
                        }
                    }
                }
            }
        }
        maxL = d[0][n];
        for(int i = m-1; i >= 0; i --) {
            for(int j = 0; j <= n; j ++) {
                int& u = d2[i][j];
                u = INF;
                if(j == 0) {
                    u = 0;
                }
                else {
                    if(i == m-1) {
                        if(p[i]/j >= maxL) {
                            u = p[i];   //else u = INF
                        }
                    }
                    else {
                        if(d[i+1][j] >= maxL) {
                            u = d2[i+1][j];
                        }
                        for(int k = 1; k <= j; k ++) {
                            if(min(p[i]/k, d[i+1][j-k]) >= maxL) {
                                u = min(u, d2[i+1][j-k] + p[i]);
                            }
                        }
                    }
                }
            }
        }
        if(!maxL) d2[0][n] = 0;
        printf("%d %d\n", d[0][n], d2[0][n]);
    }
    return 0;
}