UVA812-Trade on Verweggistan (violent), ontrade

UVA812-Trade on Verweggistan (violent), ontrade

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A merchant wants to buy pruls. Then its value is a sequence, which should be strictly obtained from start to end during purchase. For example, if you want to buy 5th, the first four should also be bought together, ask the merchant for the maximum profit.

Idea: the maximum profit must be the sum of the maximum values of each sequence. For the output, we record the position where each row can obtain the maximum value, calculate all possible values back, and then output the first 10 smallest values.

#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;const int INF = 0x3f3f3f3f;const int MAXN = 55;int arr[MAXN][MAXN], sum[MAXN][MAXN];int num[MAXN], x[MAXN * MAXN + 5];int w, b, ans, n;vector<int> v[MAXN];void dfs(int cnt, int s) {    if (cnt == w) {        x[s] = 1;           return;    }    int l = v[cnt].size();    for (int i = 0; i < l; i++)        dfs(cnt + 1, s + v[cnt][i]);}void outPut() {    int c = 10;    for (int i = 0; i < MAXN * MAXN; i++) {        if (c == 0)             break;        if (x[i]) {             printf(" %d", i);             c--;        }    }     printf("\n");}int main() {    int t = 0;    while (scanf("%d", &w) && w) {        memset(sum, 0, sizeof(sum));        memset(num, 0, sizeof(num));        for (int i = 0; i < w; i++) {            scanf("%d", &b);            num[i] = b;            for (int j = 0; j < b; j++) {                 scanf("%d", &arr[i][j]);                arr[i][j] = 10 - arr[i][j];                sum[i][j] = sum[i][j - 1] + arr[i][j];            }         }                 ans = 0;        for (int i = 0; i < w; i++) {            int Max = 0;            v[i].clear();            v[i].push_back(0);            for (int j = 0; j < num[i]; j++) {                 if (sum[i][j] > Max) {                    v[i].clear();                    v[i].push_back(j + 1);                      Max = sum[i][j];                }                else if (sum[i][j] == Max)                    v[i].push_back(j + 1);            }              ans += Max;         }         memset(x, 0, sizeof(x));        dfs(0, 0);        if (t)             printf("\n");        printf("Workyards %d\n", ++t);        printf("Maximum profit is %d.\n", ans);        printf("Number of pruls to buy:");        outPut();    }    return 0;}