题目
思路
1.状态及指标函数:d(S),S->0的最大边权路
2.状态转移方程:
3.本题NYOJ有点卡常的感觉,记忆化搜索TLE,普通递推TLE,必须把递推的顺序反以下才能AC。
代码
1.记忆化搜索
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#define _for(i,a,b) for(int i = (a); i<(b); i++)
#define _rep(i,a,b) for(int i = (a); i<=(b); i++)
using namespace std;
const int INF = 1 << 30;
const int maxn = 2000 + 10;
const int maxx = 50000 + 100;
int n, C, V[maxn], d[maxx], W[maxn];
int dp(int S) {
int &ans = d[S];
if (ans != -1) return ans;
ans = -INF;
_for(i, 0, n)
if (S >= V[i])
ans = max(ans, dp(S - V[i]) + W[i]);
return ans;
}
int main() {
int T;
scanf("%d", &T);
while (T--) {
scanf("%d%d", &n, &C);
_for(i, 0, n) scanf("%d%d", &V[i], &W[i]);
memset(d, -1, sizeof(d));
d[0] = 0;
int ans = dp(C);
if (ans <= 0) printf("NO\n");
else printf("%d\n", ans);
}
return 0;
}
2.递推
// 唯一能AC的代码
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#define _for(i,a,b) for(int i = (a); i<(b); i++)
#define _rep(i,a,b) for(int i = (a); i<=(b); i++)
using namespace std;
const int INF = 1 << 30;
const int maxn = 2000 + 10;
const int maxx = 50000 + 100;
int n, C, V[maxn], d[maxx], W[maxn];
int main() {
int T;
scanf("%d", &T);
while (T--) {
scanf("%d%d", &n, &C);
_for(i, 0, n) scanf("%d%d", &V[i], &W[i]);
_rep(i, 0, C) d[i] = -INF;
d[0] = 0;
// 注意此处把枚举的顺序反了,一般不这么做,本题是为了AC
_for(i, 0, n)
_rep(S, V[i], C)
d[S] = max(d[S], d[S - V[i]] + W[i]);
if (d[C] <= 0) printf("NO\n");
else printf("%d\n", d[C]);
}
return 0;
}