朴素做法暴力DP，O(nk)过不去。。。
1 #include <cmath> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #define N1 2005 6 #define ll long long 7 using namespace std; 8 9 int n,p; 10 int l[N1],r[N1],a[N1]; 11 ll f[N1][N1]; 12 ll linf=0x3f3f3f3f3f3fll; 13 14 int main() 15 { 16 freopen("a.txt","r",stdin); 17 scanf("%d%d",&n,&p); 18 for(int i=1;i<=n;i++) scanf("%d%d%d",&l[i],&r[i],&a[i]); 19 for(int i=0;i<=n;i++) for(int k=0;k<=p;k++) f[i][k]=linf; 20 f[0][p]=0; l[n+1]=r[n]+1; 21 for(int i=1;i<=n;i++) 22 { 23 for(int k=0;k<p;k++) 24 { 25 if(a[i]%p<=k) 26 { 27 if(a[i]/p<=r[i]-l[i]) 28 { 29 f[i][k-a[i]%p]=f[i-1][k]+a[i]/p; 30 if(a[i]/p+1<=l[i+1]-l[i]) f[i][p]=min(f[i][p],f[i-1][k]+a[i]/p+1); 31 } 32 } 33 if(a[i]%p>k) 34 { 35 if(a[i]/p+1<=r[i]-l[i]) 36 { 37 f[i][k+p-a[i]%p]=f[i-1][k]+a[i]/p+1; 38 if(a[i]/p+2<=l[i+1]-l[i]) f[i][p]=min(f[i][p],f[i-1][k]+a[i]/p+2); 39 } 40 } 41 } 42 if(a[i]%p) 43 { 44 if(a[i]/p<=r[i]-l[i]) 45 { 46 f[i][p-a[i]%p]=f[i-1][p]+a[i]/p; 47 if(a[i]/p+1<=l[i+1]-l[i]) f[i][p]=min(f[i][p],f[i-1][p]+a[i]/p+1); 48 } 49 }else{ 50 if(a[i]/p-1<=r[i]-l[i]) 51 { 52 f[i][0]=f[i-1][p]+a[i]/p-1; 53 if(a[i]/p<=l[i+1]-l[i]) f[i][p]=min(f[i][p],f[i-1][p]+a[i]/p); 54 } 55 } 56 57 } 58 ll ans=linf; 59 for(int k=0;k<=p;k++) ans=min(ans,f[n][k]*p+p-k); 60 if(ans==linf) puts("-1"); else printf("%lld ",ans); 61 return 0; 62 }
先考虑贪心这个过程
对于从i开始连续的几波，如果我们只进行不浪费子弹的换弹，我们都可以到达那些波，同时记录到达这些波换的弹夹数g[i][j]，处理出这两个数组的时间是O(n)，总时间O(n^2)
然后进行DP，f[i]记录答案，表示处理完前i-1波消耗的子弹数，那么f[i]可以由所有贪心到达i-1的点j转移过来，f[i]=min(f[j]+g[j][i-1]*p)，注意如果i和i-1之间有间隙时间，是可以转移的
总结：贪心可以处理时间充裕的情况，而DP则用来求解必须满弹开始的情况
1 #include <cmath> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #define N1 2005 6 #define ll long long 7 using namespace std; 8 9 int n,p; 10 int l[N1],r[N1],a[N1]; 11 ll f[N1],g[N1][N1],Rem[N1]; 12 bool ok[N1][N1]; 13 ll linf=0x3f3f3f3f3f3fll; 14 int divup(int x,int y) 15 { 16 if(x%y) return x/y+1; 17 else return x/y; 18 } 19 20 int main() 21 { 22 // freopen("a.txt","r",stdin); 23 scanf("%d%d",&n,&p); 24 for(int i=1;i<=n;i++) scanf("%d%d%d",&l[i],&r[i],&a[i]); 25 int rem,cnt,tim,num; 26 for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) g[i][j]=linf; 27 for(int i=2;i<=n;i++) f[i]=linf; 28 for(int i=1;i<=n;i++) 29 { 30 //greedy 31 num=divup(a[i],p); 32 if(num-1<=r[i]-l[i]) //start with full 33 { 34 g[i][i]=cnt=num, rem=num*p-a[i]; 35 Rem[i]=rem; 36 if(num<=r[i]-l[i]) //supply if empty 37 { 38 ok[i][i]=1; 39 if(rem==0) rem=p, cnt++; 40 } 41 for(int j=i+1;j<=n;j++) //never throw away remaining bullets 42 { 43 if(rem>a[j]){ 44 rem-=a[j], g[i][j]=cnt; 45 }else{ 46 num=divup(a[j]-rem,p); //never reload between consecutive waves 47 if(num<=r[j]-l[j]) g[i][j]=cnt=cnt+num, rem=num*p+rem-a[j]; 48 else{ 49 break; 50 } 51 } 52 Rem[i]=rem; 53 if(num+1<=r[j]-l[j]) 54 { 55 if(rem==0) rem=p, cnt++;//supply if empty 56 ok[i][j]=1; 57 } 58 } 59 }else{ 60 puts("-1"); return 0; 61 } 62 } 63 f[1]=0; 64 for(int i=2;i<=n;i++) 65 { 66 //DP 67 for(int j=1;j<i;j++) 68 { 69 if(g[j][i-1]!=linf && (ok[j][i-1] || r[i-1]<l[i] ) ) 70 f[i]=min(f[i],f[j]+g[j][i-1]*p); 71 } 72 } 73 ll ans=linf; 74 for(int i=1;i<=n;i++) 75 if(g[i][n]<linf/2) ans=min(ans,f[i]+g[i][n]*p-Rem[i]); 76 if(ans>=linf/2) puts("-1"); else printf("%lld ",ans); 77 return 0; 78 } 79