URAL 1987. Nested Segments（数学 & 线段树）

题目链接:http://acm.timus.ru/problem.aspx?space=1&num=1987

Input

Output

Sample

3
2 10
2 3
5 7
11
1
2
3
4
5
6
7
8
9
10
11

-1
2
2
1
3
3
3
1
1
1
-1

题意：

给定n条线段，每两条线段要么满足没有公共部分，要么包含。给出m个询问，求当前点被覆盖的最小长度的线段编号。

代码一：（先记录左边比询问点小的线段，然后再在这些点中删掉右边还比询问点小的）；

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 100017;
int tt[maxn];
int n, m;
struct node
{
    int l, r;
    int num;
} x[maxn];

bool cmp(node a, node b)
{
    return a.l < b.l;
}

int cnt = 0, k = 1;
int findd(int q)
{
    while(q >= x[k].l)
    {
        tt[++cnt] = k;
        k++;
        if(k > n)
            break;
    }
    while(q > x[tt[cnt]].r)
    {
        cnt--;
        if(cnt < 1)
            break;
    }
    return x[tt[cnt]].num;
}
int main()
{
    int ans[maxn];
    while(~scanf("%d",&n))
    {
        memset(ans, 0, sizeof(ans));
        int ll, rr;
        int minn = maxn, maxx = -1;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d%d",&ll,&rr);
            if(ll < minn)
            {
                minn = ll;
            }
            if(rr > maxx)
            {
                maxx = rr;
            }
            x[i].l = ll, x[i].r = rr;
            x[i].num = i;
        }
        int t, q[maxn];
        memset(tt, 0, sizeof(tt));
        scanf("%d",&m);
        for(int i = 1; i <= m; i++)
        {
            scanf("%d",&t);
            q[i] = t;
        }
        for(int i = 1; i <= m; i++)
        {
            if(q[i] < minn || q[i] > maxx)
            {
                ans[i] = -1;
                continue;
            }
            ans[i] = findd(q[i]);
        }
        for(int i = 1; i <= m; i++)
        {
            printf("%d\n",ans[i]);
        }
    }
    return 0;
}

由于线段不存在部分相交的情况，因此，直接按照输入顺序覆盖区间就可以了，因为后覆盖的线段更短

来自：http://www.cnblogs.com/zhsl/p/3395876.html

//STATUS:C++_AC_187MS_6805KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=100010;
const int INF=0x3f3f3f3f;
const int MOD=95041567,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

int c[(N*3)<<2],t[N][2],q[N],id[N*3];
int n;

void pushdown(int rt)
{
    if(c[rt]!=-1)
        c[rt<<1]=c[rt<<1|1]=c[rt];
}

void pushup(int rt)
{
    if(c[rt<<1]==c[rt<<1|1])
        c[rt]=c[rt<<1];
    else c[rt]=-1;
}

void update(int l,int r,int rt,int L,int R,int val)
{
    if(L<=l && r<=R){
        c[rt]=val;
        return;
    }
    pushdown(rt);
    int mid=(l+r)>>1;
    if(L<=mid)update(lson,L,R,val);
    if(R>mid)update(rson,L,R,val);
    pushup(rt);
}

int query(int l,int r,int rt,int w)
{
    if(l==r){
        return c[rt];
    }
    pushdown(rt);
    int mid=(l+r)>>1,ret;
    if(w<=mid)ret=query(lson,w);
    else ret=query(rson,w);
    pushup(rt);
    return ret;
}

int main()
{
 //   freopen("in.txt","r",stdin);
    int i,j,k,L,R,m;
    while(~scanf("%d",&n))
    {
        k=0;
        for(i=1;i<=n;i++){
            scanf("%d%d",&t[i][0],&t[i][1]);
            id[k++]=t[i][0];
            id[k++]=t[i][1];
        }
        scanf("%d",&m);
        for(i=0;i<m;i++){
            scanf("%d",&q[i]);
            id[k++]=q[i];
        }
        sort(id,id+k);
        k=unique(id,id+k)-id;
        mem(c,-1);
        for(i=1;i<=n;i++){
            L=lower_bound(id,id+k,t[i][0])-id+1;
            R=lower_bound(id,id+k,t[i][1])-id+1;
            update(1,k,1,L,R,i);
        }

        for(i=0;i<m;i++){
            printf("%d\n",query(1,k,1,lower_bound(id,id+k,q[i])-id+1));
        }
    }
    return 0;
}

URAL 1987. Nested Segments（数学 & 线段树）

原文：http://blog.csdn.net/u012860063/article/details/44256905