POJ 2184 dp

题意

传送门 POJ 2184

对于所有可能的 $TS$， $dp$ 求最大的 $FS$。实现上将负半轴映射为正值以方便数组索引。考虑 $S_{i}$ 可能取正值或负值，若 $dp$ 复用数组，则顺序应该相反。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#define min(a,b)    (((a) < (b)) ? (a) : (b))
#define max(a,b)    (((a) > (b)) ? (a) : (b))
#define abs(x)    ((x) < 0 ? -(x) : (x))
#define INF 0x3f3f3f3f
#define delta 0.85
#define eps 1e-3
#define PI 3.14159265358979323846
#define MAX_N 105
#define MAX_S 1005
using namespace std;
typedef pair<int, int> P;
int N;
P cow[MAX_N];
int dp[2 * MAX_S * MAX_N];

int main(){
	while(~scanf("%d", &N)){
		memset(dp, 0x3f, sizeof(dp));
		int mid = MAX_S * N;
		for(int i = 0; i < N; i++) scanf("%d%d", &cow[i].first, &cow[i].second);
		dp[mid] = 0;
		for(int i = 0; i < N; i++){
			int limit = MAX_S * (i + 1);
			P &c = cow[i];
			// Si 为正值
			if(c.first >= 0){
				for(int j = limit + mid; j >= -limit + mid; j--){
					if(dp[j] != INF){
						int nxt = j + c.first, sum = c.second + dp[j];
						dp[nxt] = dp[nxt] == INF ? sum : max(sum, dp[nxt]);
					}
				}
			}
			// Si 为负值
			else{
				for(int j = -limit + mid; j <= limit + mid; j++){
					if(dp[j] != INF){
						int nxt = j + c.first, sum = c.second + dp[j];
						dp[nxt] = dp[nxt] == INF ? sum : max(sum, dp[nxt]);
					}
				}
			}
			
		}
		int res = 0;
		for(int i = mid; i <= mid + MAX_S * N; i++){
			if(dp[i] != INF && dp[i] >= 0) res = max(res, dp[i] + i - mid);
		}
		printf("%d\n", res);
	}
	return 0;
}