UVA - 812 Trade on Verweggistan dfs ＋ 贪心

题目大意：有n堆东西，只能从上往下买，如果要买下面的东西，只能把上面的所有东西买了才可以，类似一个栈
每个东西都有价格，每个东西卖出去都只能卖10，问能卖的最大利润，和要卖多少个才能达到最大利润

解题思路：贪心的来说每个买了的都要卖出去，才能使得利润最大化，纪录每堆卖出最大利润的所有情况，然后用dfs一一枚举，要判重和排序输出
注意特例，都是0的情况
1
5 10 10 10 10 10
输出应该有0 1 2 3 4 5

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define maxn 60int w, price[maxn][25], sum, num[maxn], cnt, vis[2500],ans[2500];struct Max {    int cnt;    int M[25];    int num;}MAX[maxn];void init() {    int Max = 0, sum2;    sum = 0; cnt = 0;    memset(vis,0,sizeof(vis));    for(int i = 0; i < w; i++) {        scanf("%d", &num[i]);        sum += Max;        sum2 = Max = 0;        for(int j = 0; j < num[i]; j++) {            scanf("%d", &price[i][j]);            sum2 += 10 - price[i][j];            Max = max(Max, sum2);        }        MAX[i].num = Max;        MAX[i].cnt = 0;    }    sum += Max;    for(int i = 0; i < w; i++) {        sum2 = 0;        if(MAX[i].num == 0)            MAX[i].M[MAX[i].cnt++] = 0;        for(int j = 0; j < num[i]; j++) {            sum2 += 10 -price[i][j];            if(sum2 == MAX[i].num)                MAX[i].M[MAX[i].cnt++] = j + 1;        }    }}void dfs(int Sum, int cur) {    if(cur == w && !vis[Sum]) {        ans[cnt++] =  Sum;        vis[Sum] = 1;        return ;    }    for(int i = 0; i < MAX[cur].cnt; i++) {        Sum += MAX[cur].M[i];        dfs(Sum,cur+1);        Sum -= MAX[cur].M[i];    }}int main() {    int cas = 0;    while(scanf("%d", &w) == 1 && w) {        if(cas)            printf("\n");        init();        dfs(0,0);        sort(ans,ans+cnt);        printf("Workyards %d\n", ++cas);        printf("Maximum profit is %d.\n", sum);        printf("Number of pruls to buy:");                for(int i = 0; i < cnt && i < 10; i++)            printf(" %d", ans[i]);        printf("\n");    }    return 0;}