Description
Input
Output
对于每组数据，若最小的不协调度不超过1018，则第一行一个数表示不协调度若最小的不协调度超过1018，则输出"Too hard to arrange"(不包含引号)。每个输出后面加"--------------------"
Sample Input
4
4 9 3
brysj,
hhrhl.
yqqlm,
gsycl.
4 9 2
brysj,
hhrhl.
yqqlm,
gsycl.
1 1005 6
poet
1 1004 6
poet
4 9 3
brysj,
hhrhl.
yqqlm,
gsycl.
4 9 2
brysj,
hhrhl.
yqqlm,
gsycl.
1 1005 6
poet
1 1004 6
poet
Sample Output
108
--------------------
32
--------------------
Too hard to arrange
--------------------
1000000000000000000
--------------------
--------------------
32
--------------------
Too hard to arrange
--------------------
1000000000000000000
--------------------
【样例说明】
前两组输入数据中每行的实际长度均为6，后两组输入数据每行的实际长度均为4。一个排版方案中每行相邻两个句子之间的空格也算在这行的长度中(可参见样例中第二组数据)。每行末尾没有空格。
HINT
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。
题解如下
https://www.byvoid.com/blog/noi-2009-poet
我只放AC代码
#include<cstdio>
#include<cstring>
#define ll long double
struct node{int l,r,p;}q[];
#define MAX 1000000000000000000LL
#define N 100100
ll sum[N],f[N];
int n,l,p,T;
char ch[];
ll pow(ll y){
if(y<)y=-y;
ll ans=;
for (int i=;i<=p;i++) ans*=y;
return ans;
} ll calc(int x,int y){
return f[x]+pow(sum[y]-sum[x]+(y-x-)-l);
} int find(node t,int x){
int l=t.l,r=t.r;
while(l<=r){
int mid=(l+r)>>;
if (calc(x,mid)<=calc(t.p,mid)) r=mid-;
else l=mid+;
}
return l;
} void dp(){
int head=,tail=;
q[++tail]=(node){,n,};
for (int i=;i<=n;i++){
if(q[head].r<i&&head<=tail) head++;
f[i]=calc(q[head].p,i);
if (head>tail||calc(i,n)<=calc(q[tail].p,n)){
while(head<=tail&&calc(q[tail].p,q[tail].l)>=calc(i,q[tail].l)) tail--;
if(head>tail)q[++tail]=(node){i,n,i};
else{
int x=find(q[tail],i);
q[tail].r=x-;
q[++tail]=(node){x,n,i};
}
}
}
} int main(){
scanf("%d",&T);
while(T--){
scanf("%d%d%d",&n,&l,&p);
for (int i=;i<=n;i++) scanf("%s",ch),sum[i]=sum[i-]+strlen(ch);
dp();
if(f[n]>MAX)
puts("Too hard to arrange");
else
printf("%lld\n",(long long)f[n]);
puts("--------------------");
}
}