FatMouse's Speed
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6632 Accepted Submission(s): 2898
Special Judge
The data for a particular mouse will consist of a pair of integers: the first representing its size in grams and the second representing its speed in centimeters per second. Both integers are between 1 and 10000. The data in each test case will contain information for at most 1000 mice.
Two mice may have the same weight, the same speed, or even the same weight and speed.
W[m[1]] < W[m[2]] < ... < W[m[n]]
and
S[m[1]] > S[m[2]] > ... > S[m[n]]
In order for the answer to be correct, n should be as large as possible.
All inequalities are strict: weights must be strictly increasing, and speeds must be strictly decreasing. There may be many correct outputs for a given input, your program only needs to find one.
這是一道DP的題目,之前已經上了好幾題DP的了。。。繼續。
解題思路:
對W或S排序後,對另一個進行DP。
(未完待續,下次上圖說明)
代碼:
#include <stdio.h>
int main(int argc, const char * argv[])
{
int a[1005],b[1005],c[1005],d[1005],i,j,k,t,m=0,w,v[1005]={-1};
i=0;
while(scanf("%d",&a[i])!=EOF/*&&a[i]!=0*/)
{
scanf("%d",&b[i++]);
}
//sort
for(j=0;j<i;j++)
{
t=0;
for(w=0;w<i;w++)
{
if(a[w]>t)
{
t=a[w];
k=w;
}
}
c[m++]=k;
a[k]=0;
}
//find max
for(j=0;j<i;j++)
{
k=c[j];
w=j;
if(j==0) {d[k]=1;v[k]=-1;}
else{
m=1;
for(t=c[w-1];w>=0;w--)
{
t=c[w-1];
if(b[k]>b[t])
{if(d[t]>=m) {m=d[t]+1;v[k]=t;}}
}
d[k]=m;if(m==1) v[k]=-1;
}
}
// output
t=0;
for(j=0;j<i;j++)
{
k=c[j];
if(d[k]>t) {t=d[k];w=c[j];}
}
printf("%d\n",t);
//print
printf("%d\n",w+1);
while(d[w]!=1){
w=v[w];
if(w==-1) break;
printf("%d\n",w+1);
}
return 0;
}