UVa #10163 Storage Keepers (习题9-9)

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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 Keeperschar 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;}

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