[二分][状压dp] Jzoj P3521 道路覆盖
题解
- 最小值最大,显然二分
- 现在二分出来一个最低高度,思考一下怎么判断
- 考虑dp,设f[i][s]为当前到第i位前k个位用泥土的状态为s(是否有用)
- 那么就很显然了,每次转移前现算出前k个给当前位的贡献,设为num
- 转移就有两种一种是当前位不选,一种是当前位选
- 状态转移方程就很显然了
- ①f[i+1][j>>1]=min(f[i+1][j>>1],f[i][j]);
②f[i+1][(j>>1)|(1<<(k-1))]=min(f[i+1][(j>>1)|(1<<(k-1))],f[i][j]+c[i+1]);
代码
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 using namespace std; 5 int n,m,k,sum,mx,num,h[110],e[110],c[110],f[110][1<<12]; 6 bool check(int x) 7 { 8 memset(f,127,sizeof(f)),f[0][0]=0; 9 for (int i=0;i<=n-1;i++) 10 for (int j=0;j<=(1<<k)-1;j++) 11 { 12 num=0; 13 for (int z=1;z<=k-1;z++) if (j&(1<<z)) num+=e[i-k+z+1]; 14 if (num+h[i+1]>=x) f[i+1][j>>1]=min(f[i+1][j>>1],f[i][j]); 15 if (num+h[i+1]+e[i+1]>=x) f[i+1][(j>>1)|(1<<(k-1))]=min(f[i+1][(j>>1)|(1<<(k-1))],f[i][j]+c[i+1]); 16 } 17 for (int i=0;i<=(1<<k)-1;i++) if (f[n][i]<=m) return true; 18 return false; 19 } 20 int main() 21 { 22 freopen("cover.in","r",stdin),freopen("cover.out","w",stdout); 23 scanf("%d%d%d",&n,&m,&k); 24 for (int i=1;i<=n;i++) scanf("%d%d%d",&h[i],&e[i],&c[i]),sum+=e[i],mx=max(mx,h[i]); 25 int l=1,r=sum+mx; 26 while (l<r) 27 { 28 int mid=(l+r)>>1; 29 if (check(mid)) l=mid+1; else r=mid; 30 } 31 printf("%d",l-1); 32 }