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总共10个测试点,数据范围满足:
测试点 T N L P
1 ≤10 ≤18 ≤100 ≤5
2 ≤10 ≤2000 ≤60000 ≤10
3 ≤10 ≤2000 ≤60000 ≤10
4 ≤5 ≤100000 ≤200 ≤10
5 ≤5 ≤100000 ≤200 ≤10
6 ≤5 ≤100000 ≤3000000 2
7 ≤5 ≤100000 ≤3000000 2
8 ≤5 ≤100000 ≤3000000 ≤10
9 ≤5 ≤100000 ≤3000000 ≤10
10 ≤5 ≤100000 ≤3000000 ≤10
所有测试点中均满足句子长度不超过30。
1 #include<cstdio> 2 #include<cstring> 3 #define ll long double 4 struct node{int l,r,p;}q[100100]; 5 #define MAX 1000000000000000000LL 6 #define N 100100 7 ll sum[N],f[N]; 8 int n,l,p,T; 9 char ch[35]; 10 ll pow(ll y){ 11 if(y<0)y=-y; 12 ll ans=1; 13 for (int i=1;i<=p;i++) ans*=y; 14 return ans; 15 } 16 17 ll calc(int x,int y){ 18 return f[x]+pow(sum[y]-sum[x]+(y-x-1)-l); 19 } 20 21 int find(node t,int x){ 22 int l=t.l,r=t.r; 23 while(l<=r){ 24 int mid=(l+r)>>1; 25 if (calc(x,mid)<=calc(t.p,mid)) r=mid-1; 26 else l=mid+1; 27 } 28 return l; 29 } 30 31 void dp(){ 32 int head=1,tail=0; 33 q[++tail]=(node){0,n,0}; 34 for (int i=1;i<=n;i++){ 35 if(q[head].r<i&&head<=tail) head++; 36 f[i]=calc(q[head].p,i); 37 if (head>tail||calc(i,n)<=calc(q[tail].p,n)){ 38 while(head<=tail&&calc(q[tail].p,q[tail].l)>=calc(i,q[tail].l)) tail--; 39 if(head>tail)q[++tail]=(node){i,n,i}; 40 else{ 41 int x=find(q[tail],i); 42 q[tail].r=x-1; 43 q[++tail]=(node){x,n,i}; 44 } 45 } 46 } 47 } 48 49 int main(){ 50 scanf("%d",&T); 51 while(T--){ 52 scanf("%d%d%d",&n,&l,&p); 53 for (int i=1;i<=n;i++) scanf("%s",ch),sum[i]=sum[i-1]+strlen(ch); 54 dp(); 55 if(f[n]>MAX) 56 puts("Too hard to arrange"); 57 else 58 printf("%lld\n",(long long)f[n]); 59 puts("--------------------"); 60 } 61 }
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原文地址:http://www.cnblogs.com/wuminyan/p/5191639.html