先展示一个我同学些的代码，超级good
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#include<iostream> #include<cstdio> #include<cstring> using namespace std; int C[150005]; int B[150005]; //int Out[150005]; int N=150005; int Lowbit(int x) { return x&(-x); } void Modify(int i,int x) { while(i>0) { cout<<i<<" : "<<C[i]<<" "<<Lowbit(i)<<endl; C[i]+=x; i-=Lowbit(i); } cout<<i<<" : "<<C[i]<<endl; } int Sum(int n) { int sum=0; while(n<=N) { cout<<n<<" "<<C[n]<<" "<<Lowbit(n)<<endl; sum+=C[n]; n+=Lowbit(n); } return sum; } int main() { int t; cin>>t; int Case=1; while(t--) { int i,n,m; int x,y; memset(C,0,sizeof(C)); memset(B,0,sizeof(B)); // memset(Out,0,sizeof(Out)); scanf("%d%d",&n,&m); for (i=1;i<=n;i++) { scanf("%d%d",&x,&y); Modify(x-1,-1); Modify(y,1); // Out[Sum(x)-1]++; } printf("Case #%d:\n",Case); int q; for (i=1;i<=m;i++) { scanf("%d",&q); printf("%d\n",Sum(q)); } Case++; } return 0; }
为了节约时间减少数据量的算法，这才是算法嘛！这个是网上找的，很复杂，勉强看懂，保存！
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#include<iostream> #include<cstdio> #include<algorithm> using namespace std; #define N 100005 struct node{ int l,r,s; }a[N*6]; int pa[N*2],cou,sum,cou1,tt; int aa[N],bb[N],qq[N]; void build(int i,int left,int right){ //i=1 a[i].l=left; a[i].r=right; a[i].s=0; cout<<" build a["<<i<<"]:"<<a[i].l<<' '<<a[i].r<<' '<<a[i].s<<endl; if(a[i].l==a[i].r) return ; int mid=(left+right)>>1; //相当于除以 2 build(i*2,left,mid); //i=2 4 5 8 9 10 11 .... build(i*2+1,mid+1,right); //i=3 6 7 12 13 14 15 .... } void insert(int i,int left,int right){ if(a[i].l==left&&a[i].r==right){ a[i].s++; cout<<"insert a["<<i<<"]:"<<a[i].l<<' '<<a[i].r<<' '<<a[i].s<<endl; return ; } int mid=(a[i].l+a[i].r)>>1; if(left>mid) insert(i*2+1,left,right); else if(right<=mid) insert(i*2,left,right); else{ insert(i*2,left,mid); insert(i*2+1,mid+1,right); } } int querry(int i,int x){ sum+=a[i].s; if(x==a[i].l&&x==a[i].r){ return sum; } int mid=(a[i].l+a[i].r)>>1; if(x>mid) { querry(i*2+1,x); } else if(x<=mid) { querry(i*2,x); } } int find(int x){ int l=1,r=tt-1; while(r>=l){ int mid=(l+r)>>1; if(x>pa[mid]) l=mid+1; if(x<pa[mid]) r=mid-1; if(x==pa[mid]) return mid; } return 0; } int main(){ // freopen("in.txt","r",stdin); int t,n,m,x,y,z,p=1; scanf("%d",&t); while(t--){ scanf("%d%d",&n,&m); cou=1; for(int i=0;i<n;i++){ scanf("%d%d",&x,&y); aa[i]=x; bb[i]=y; pa[cou++]=x; pa[cou++]=y; } for(int i=0;i<m;i++){ scanf("%d",&qq[i]); pa[cou++]=qq[i]; } printf("Case #%d:\n",p++); sort(pa+1,pa+cou); for(int i=1;i<cou;i++) cout<<pa[i]<<" "; cout<<endl; int cou1=1; pa[cou]=-1; for(int i=1;i<cou;i++){ if(pa[i]!=pa[i+1]) pa[cou1++]=pa[i]; else continue; } for(int i=1;i<cou1;i++) cout<<pa[i]<<" "; cout<<endl; build(1,0,cou1-1); tt=cou1; for(int i=0;i<n;i++){ int ta=find(aa[i]); int tb=find(bb[i]); insert(1,ta,tb); } for(int i=0;i<m;i++){ int z=find(qq[i]); sum=0; printf("%d\n",querry(1,z)); } } return 0; }
下面是我简单的头脑可以想出来的。不明白为什么同样的方法用c++写超时了而c写就过了
Flowers
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 288 Accepted Submission(s): 194
Problem Description
As is known to all, the blooming time and duration varies between different kinds of flowers. Now there is a garden planted full of flowers. The gardener wants to know how many flowers will bloom in the garden in a specific time. But there are too many flowers in the garden, so he wants you to help him.
Input
The first line contains a single integer t (1 <= t <= 10), the number of test cases.
For each case, the first line contains two integer N and M, where N (1 <= N <= 10^5) is the number of flowers, and M (1 <= M <= 10^5) is the query times.
In the next N lines, each line contains two integer Si and Ti (1 <= Si <= Ti <= 10^9), means i-th flower will be blooming at time [Si, Ti].
In the next M lines, each line contains an integer Ti, means the time of i-th query.
Output
For each case, output the case number as shown and then print M lines. Each line contains an integer, meaning the number of blooming flowers.
Sample outputs are available for more details.
Sample Input
2
1 1
5 10
4
2 3
1 4
4 8
1
4
6
Sample Output
Case #1:
0
Case #2:
1
2
1
这个超时了
#include<iostream> #include<string.h> using namespace std; int p[10000000]; int main( ) { int T,i; cin>>T; for(int Case=1; Case<=T; ++ Case) { cout<<"Case #"<<Case<<":"<<endl; int N, M,a, b; cin>>N>>M; memset( p,0,sizeof p); for( i=0;i<N;++i ) { cin>>a>>b; for( int j=a; j<=b; ++j ) p[j]++; } for( i=0;i<M;++i ) { int x; cin>>x; cout<<p[x]<<endl; } } return 0; }
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2012-08-01 08:57:20 | Time Limit Exceeded | 4325 | 2000MS | 39468K | 579 B | C++ |
下面是c写的代码。这个就AC了。希望哪位看到的大神指点指点。。。
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#include <stdio.h> #include <string.h> int p[10000000]; int main( ) { int T,i; scanf( "%d", &T ); for(int Case=1; Case<=T; ++ Case) { printf( "Case #%d:\n",Case ); int N, M,a, b; scanf("%d%d", &N,&M ); memset( p,0,sizeof p); for( i=0;i<N;++i ) { scanf( "%d%d", &a, &b ); for( int j=a; j<=b; ++j ) p[j]++; } for( i=0;i<M;++i ) { int x; scanf( "%d", &x ); printf("%d\n", p[x] ); } } return 0; }
| Run ID | Submit Time | Judge Status | Pro.ID | Exe.Time | Exe.Memory | Code Len. | Language |
| 2012-08-01 08:57:50 | Accepted | 4325 | 1656MS | 39432K | 610 B | C++ |
