(problem)
此题是一道01背包。
关于01 背包 我不想讲了 - > (MY BLOG)
[这道题是一道基础的01背包问题
]
[设f[i][j]=k表示前i张牌构成分值j的最小次数k
]
[设 dis = a[i]-b[i]
]
//不反转
[f[i][j+dis+N]=min(f[i][j+dis+N],f[i-1][j+N]);
]
//反转
[f[i][j+dis+N]=min(f[i][j+dis+N],f[i-1][j+N]+1);
]
所以得出
[f[i][j+N] = min(f[i-1][j-dis+N] , f[i-1][j+dis+N] + 1) ;
]
#include <bits/stdc++.h>
#define rep(i,j,n) for(register int i=j;i<=n;i++)
#define Rep(i,j,n) for(register int i=j;i>=n;i--)
#define low(x) x&(-x)
using namespace std ;
typedef long long LL ;
const int inf = INT_MAX >> 1 ;
inline LL In() { LL res(0) , f(1) ; register char c ;
#define gc c = getchar()
while(isspace(gc)) ; c == '-' ? f = - 1 , gc : 0 ;
while(res = (res << 1) + (res << 3) + (c & 15) , isdigit(gc)) ;
return res * f ;
#undef gc
}
int n ;
const int Size = 1000 + 5 ;
const int N = 5000 ;
int a[Size] , b[Size] ;
int f[Size][Size * 11] ;
int ans ;
inline void Ot() {
n = In() ;
rep(i,1,n) a[i] = In() , b[i] = In() ;
memset(f,0x7f,sizeof(f)) ;
f[0][N] = 0 ;
rep(i,1,n) rep(j,-N,N) {
int dis = a[i] - b[i] ;
f[i][j+N] = min(f[i-1][j-dis+N] , f[i-1][j+dis+N] + 1) ;
}
rep(i,0,N) {
int ans = min(f[n][N+i] , f[n][N-i]) ;
if(ans <= 1000) {
printf("%d
",ans) ;
return ;
}
}
}
signed main() {
// freopen("testdata.txt","w",stdout) ;
return Ot() , 0 ;
}