UVA 11572 Unique Snowflakes .
2016-11-10 09:15
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题目地址:http://vjudge.net/problem/UVA-11572
第一种用set做
#include <bits/stdc++.h>
using namespace std;
#define REP(i,a,b) for(int i=a;i<=(int)(b);++i)
#define REPD(i,a,b) for(int i=a;i>=(int)(b);--i)
int A[(int)1e6+5];
int main(int argc, char const *argv[])
{
int T,n; scanf("%d",&T);
while(T--&&scanf("%d",&n)==1){
REP(i,1,n) scanf("%d",&A[i]);
set<int> st;
int L=1,R=1,ans=1;
while(R<=n){
while(R<=n&&!st.count(A[R])) st.insert(A[R]),R++;
ans=max(ans,R-L);
st.erase(A[L]); L++;
}
printf("%d\n", ans);
}
return 0;
}
第二种,用last保存上一个相同数字的坐标 来代替set查找
#include <bits/stdc++.h>
using namespace std;
#define REP(i,a,b) for(int i=a;i<=(int)(b);++i)
#define REPD(i,a,b) for(int i=a;i>=(int)(b);--i)
int A[1000000+5],last[1000000+5];
int main(int argc, char const *argv[])
{
int T,n; scanf("%d",&T);
while(T--&&scanf("%d",&n)==1){
map<int,int> cur;
REP(i,1,n) {
scanf("%d",&A[i]);
if(cur.count(A[i])==0) last[i]=-1;
else last[i]=cur[A[i]];
cur[A[i]]=i;
}
int L=1,R=1,ans=1;
while(R<=n){
while(R<=n&&last[R]<L) R++;
ans=max(ans,R-L);
L++;
}
printf("%d\n", ans);
}
return 0;
}
第一种用set做
#include <bits/stdc++.h>
using namespace std;
#define REP(i,a,b) for(int i=a;i<=(int)(b);++i)
#define REPD(i,a,b) for(int i=a;i>=(int)(b);--i)
int A[(int)1e6+5];
int main(int argc, char const *argv[])
{
int T,n; scanf("%d",&T);
while(T--&&scanf("%d",&n)==1){
REP(i,1,n) scanf("%d",&A[i]);
set<int> st;
int L=1,R=1,ans=1;
while(R<=n){
while(R<=n&&!st.count(A[R])) st.insert(A[R]),R++;
ans=max(ans,R-L);
st.erase(A[L]); L++;
}
printf("%d\n", ans);
}
return 0;
}
第二种,用last保存上一个相同数字的坐标 来代替set查找
#include <bits/stdc++.h>
using namespace std;
#define REP(i,a,b) for(int i=a;i<=(int)(b);++i)
#define REPD(i,a,b) for(int i=a;i>=(int)(b);--i)
int A[1000000+5],last[1000000+5];
int main(int argc, char const *argv[])
{
int T,n; scanf("%d",&T);
while(T--&&scanf("%d",&n)==1){
map<int,int> cur;
REP(i,1,n) {
scanf("%d",&A[i]);
if(cur.count(A[i])==0) last[i]=-1;
else last[i]=cur[A[i]];
cur[A[i]]=i;
}
int L=1,R=1,ans=1;
while(R<=n){
while(R<=n&&last[R]<L) R++;
ans=max(ans,R-L);
L++;
}
printf("%d\n", ans);
}
return 0;
}
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