poj1010

dfs

对于多解的判断很复杂。另外注意输出的括号中的数字是邮票的种类。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;

#define maxn 300

int n;
int m, tot;
int stamp[maxn];
int ans[5], ansnum;
int sel[5];
bool tie, none;
bool vis[maxn];

void input()
{
    n = 1;
    if (scanf("%d", &stamp[0]) == EOF)
        exit(0);
    while (scanf("%d", &stamp[n]), stamp[n])
        n++;
}

int cal(int a[], int n)
{
    int ret = 0;
    memset(vis, 0, sizeof(vis));
    for (int i = 0; i < n; i++)
        if (!vis[a[i]])
        {
            vis[a[i]] = true;
            ret++;
        }
    return ret;
}

int getmax(int a[], int n)
{
    int ret = 0;
    for (int i = 0; i < n; i++)
        ret = max(ret, stamp[a[i]]);
    return ret;
}

void compare()
{
    int ksel = cal(sel, tot);
    int kans = cal(ans, ansnum);
    int maxans = getmax(ans, ansnum);
    int maxsel = getmax(sel, tot);

    if (ansnum == -1 || ksel > kans || (ksel == kans && ansnum > tot) || (ksel
            == kans && ansnum == tot && maxans < maxsel))
    {
        tie = false;
        ansnum = tot;
        for (int i = 0; i < tot; i++)
            ans[i] = sel[i];
        return;
    }
    if (ksel == kans && ansnum == tot && maxans == maxsel)
        tie = true;
}

void dfs(int now, int money)
{
    if (money > m)
        return;
    if (money == m)
    {
        none = false;
        compare();
    }
    if (tot == 4)
        return;
    for (int i = now; i < n; i++)
    {
        sel[tot] = i;
        tot++;
        dfs(i, money + stamp[i]);
        tot--;
    }
}

void print()
{
    if (none)
    {
        printf("%d ---- none\n", m);
        return;
    }
    printf("%d (%d):", m, cal(ans, ansnum));
    if (tie)
    {
        printf(" tie\n");
        return;
    }
    for (int i = 0; i < ansnum; i++)
        printf(" %d", stamp[ans[i]]);
    putchar('\n');
}

int main()
{
    //freopen("t.txt", "r", stdin);
    while (1)
    {
        input();
        sort(stamp, stamp + n);
        while (scanf("%d", &m), m)
        {
            tot = 0;
            ansnum = -1;
            tie = false;
            none = true;
            dfs(0, 0);
            if (!none)
                sort(ans, ans + ansnum);
            print();
        }
    }
    return 0;
}