ZOJ - 3623 完全背包变种

题意理解有误导致方程建歪,题意是n种类型的船造成至少L伤害的最小时间,攻击过程是不必同步的

#include<iostream>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cmath>

#include<string>

#include<vector>

#include<stack>

#include<queue>

#include<set>

#include<map>

#define rep(i,j,k) for(register int i=j;i<=k;i++)

#define rrep(i,j,k) for(register int i=j;i>=k;i--)

#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])

#define iin(a) scanf("%d",&a)

#define lin(a) scanf("%lld",&a)

#define din(a) scanf("%lf",&a)

#define s0(a) scanf("%s",a)

#define s1(a) scanf("%s",a+1)

#define print(a) printf("%lld",(ll)a)

#define enter putchar('\n')

#define blank putchar(' ')

#define println(a) printf("%lld\n",(ll)a)

#define IOS ios::sync_with_stdio(0)

using namespace std;

const int maxn = 1e6+11;

const int oo = 0x3f3f3f3f;

const double eps = 1e-7;

typedef long long ll;

ll read(){

    ll x=0,f=1;register char ch=getchar();

    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

    return x*f;

}

int n,L,cost[maxn],weight[maxn],tot;

int dp[maxn];

int main(){

	while(cin>>n>>L){

		rep(i,1,n){

			cost[i]=read();

			weight[i]=read();

		}

		memset(dp,0,sizeof dp);

		int ans=oo;

		//时间为j时能生产船伤害的最大总值

//		rep(i,1,n){

//			rep(j,cost[i],n*L){

//				dp[j]=max(dp[j],dp[j-cost[i]]+weight[i]);

//				if(dp[j]==0) continue;

//				else ans=min(ans,j+(int)ceil(1.0*L/dp[j]));//这个是必须建好后一起攻击

//			}

//		}

		rep(i,1,n){

			rep(j,cost[i],6667){

				dp[j]=max(dp[j],dp[j-cost[i]]+(j-cost[i])*weight[i]);//这个是一旦建好船久可以直接攻击

			}

		}

		rep(i,1,6667) if(dp[i]>=L) ans=min(ans,i);

		println(ans);

	}

	return 0;

}