fzu 1656 How many different numbers(线段树)

题意：询问不同区间不同值的个数

//思路1 700+ms#include<iostream>#include<string>#include<algorithm>using namespace std;#define CL(a,b) memset(a,b,sizeof(a))const int M(100010);const int N(1020);int f[N],r[M];struct data{int x,sub;}a[M];struct dat{int l,r;int sub;}ans[N];int vis[M];bool cmp(data x,data y){return x.x<y.x;}bool cm(dat x,dat y ){return x.l<y.l||(x.l==y.l&&x.r<y.r);}int main(){int m,i,n,j,q;while(scanf("%d",&m)!=EOF){for(i=1;i<=m;i++){scanf("%d",&a[i].x);a[i].sub=i;}sort(a+1,a+m+1,cmp);r[a[1].sub]=0;for(i=2,j=0;i<=m;i++)//离散化{if(a[i-1].x==a[i].x)r[a[i].sub]=j;else{j++;r[a[i].sub]=j;}}scanf("%d",&q);for(i=0;i<q;i++){scanf("%d%d",&ans[i].l,&ans[i].r);ans[i].sub=i;}sort(ans,ans+q,cm);CL(f,0); CL(vis,0);//vis记录出现个数for(i=0;i<q;i++){if(i==0){CL(vis,0);for(j=ans[i].l;j<=ans[i].r;j++){if(!vis[r[j]])f[ans[i].sub]++;vis[r[j]]++;}continue;}if(ans[i].r<ans[i-1].r){if((ans[i].l-ans[i-1].l+ans[i-1].r-ans[i].r)>(ans[i].r-ans[i].l)){CL(vis,0);for(j=ans[i].l;j<=ans[i].r;j++){if(!vis[r[j]])f[ans[i].sub]++;vis[r[j]]++;}}else{f[ans[i].sub]=f[ans[i-1].sub];for(j=ans[i-1].l;j<ans[i].l;j++){vis[r[j]]--;if(vis[r[j]]==0)f[ans[i].sub]--;}for(j=ans[i].r+1;j<=ans[i-1].r;j++){vis[r[j]]--;if(vis[r[j]]==0)f[ans[i].sub]--;}}}else {if((ans[i].l-ans[i-1].l)>(ans[i-1].r-ans[i].l)){CL(vis,0);for(j=ans[i].l;j<=ans[i].r;j++){if(!vis[r[j]])f[ans[i].sub]++;vis[r[j]]++;}}else{f[ans[i].sub]=f[ans[i-1].sub];for(j=ans[i-1].l;j<ans[i].l;j++){vis[r[j]]--;if(vis[r[j]]==0)f[ans[i].sub]--;}for(j=ans[i-1].r+1;j<=ans[i].r;j++){if(!vis[r[j]])f[ans[i].sub]++;vis[r[j]]++;}}}}for(i=0;i<q;i++)printf("%d\n",f[i]);}}

//参考：http://blog.csdn.net/sqplfh/article/details/6840996

具体思路：先把N个保存下来，进行离散，然后用就可以用flag记录某个数前面出现的位置，再把询问保存下来，按右端点排序。

对N个数依次处理，删除该数前面出现的，在当前位置插入该数。当处理到第i个位置的数，且询问的右端点也为i是，统计个数。

// 300+ms 线段树优化#include<iostream>#include<string>#include<algorithm>using namespace std;#define CL(a,b) memset(a,b,sizeof(a))#define MID(a,b) (a+b)>>1const int M(100010);const int N(1020);int tree[M<<2],r[M],f[N];struct data{int x,sub;}a[M];struct dat{int l,r;int sub;}ans[N];int vis[M];bool cmp(data x,data y){return x.x<y.x;}bool cm(dat x,dat y ){return x.r<y.r;}inline void pushup(int t){tree[t]=tree[t<<1]+tree[t<<1|1];}void insert(int t,int x,int val,int l,int r){if(l==r){tree[t]=val;return;}int mid=MID(l,r);if(x<=mid)insert(t<<1,x,val,l,mid);if(x>mid)insert(t<<1|1,x,val,mid+1,r);pushup(t);}int query(int t,int ql,int qr,int l,int r){if(ql<=l&&qr>=r)return tree[t];int mid=MID(l,r);if(qr<=mid)return query(t<<1,ql,qr,l,mid);else if(ql>mid)return query(t<<1|1,ql,qr,mid+1,r);elsereturn query(t<<1,ql,qr,l,mid)+query(t<<1|1,ql,qr,mid+1,r);}int main(){int m,i,n,j,q;while(scanf("%d",&m)!=EOF){for(i=1;i<=m;i++){scanf("%d",&a[i].x);a[i].sub=i;}sort(a+1,a+m+1,cmp);r[a[1].sub]=0;for(i=2,j=0;i<=m;i++){if(a[i-1].x==a[i].x)r[a[i].sub]=j;else{j++;r[a[i].sub]=j;}}scanf("%d",&q);for(i=0;i<q;i++){scanf("%d%d",&ans[i].l,&ans[i].r);ans[i].sub=i;}sort(ans,ans+q,cm);CL(vis,-1); CL(tree,0);for(i=1,j=0;i<=m&&j<q;i++){if(vis[r[i]]!=-1){insert(1,vis[r[i]],0,1,m);}insert(1,i,1,1,m);vis[r[i]]=i;while(i==ans[j].r){f[ans[j].sub]=query(1,ans[j].l,ans[j].r,1,m);j++;//printf("%d\n",tree[1]);}}for(i=0;i<q;i++)printf("%d\n",f[i]);}}