PKU 2184 - cow Exhibition (0-1背包)
题意
把第一个条件当成重量,第二个条件当成权值,就变成01背包了。
处理一下负数。
代码
#include <stack>
#include <cstdio>
#include <list>
#include <cassert>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define X first
#define Y second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int MAXN = 200000 + 10;
const int MOD = 9901;
const int dir[][2] = { {-1, 0}, {0, -1}, { 1, 0 }, { 0, 1 } };
int cases = 0;
typedef pair<int, int> pii;
int dp[MAXN];
pii arr[110];
int main()
{
//ROP;
int n, k = 100000, kk = 200000;
scanf("%d", &n);
for (int i = 1; i <= n; i++)
scanf("%d%d", &arr[i].X, &arr[i].Y);
for (int i = 0; i <= kk; i++) dp[i] = -INF;
dp[k] = 0;
for (int i = 1; i <= n; i++)
{
if (arr[i].X > 0)
for (int j = kk; j >= arr[i].X; j--)
dp[j] = max(dp[j], dp[j-arr[i].X]+arr[i].Y);
else
for (int j = 0; j <= kk+arr[i].X; j++)
dp[j] = max(dp[j], dp[j-arr[i].X]+arr[i].Y);
}
int ans = 0;
for (int i = k; i <= kk; i++)
if (dp[i] > 0) ans = max(ans, dp[i]+i-k);
printf("%d\n", ans);
return 0;
}