2017 ACM-ICPC 亚洲区（南宁赛区）网络赛 (B,F,L,M)

B. Train Seats Reservation

You are given a list of train stations, say from the station

The passengers can order several tickets from one station to another before the train leaves the station one. We will issue one train from the station

Note that one single seat can be used by several passengers as long as there are no conflicts between them. For example, a passenger from station

Input Format

Several sets of ticket reservations. The inputs are a list of integers. Within each set, the first integer (in a single line) represents the number of orders,

Output Format

For each set of ticket reservations appeared in the input, calculate the minimum number of seats required so that all reservations are satisfied without conflicts. Output a single star '*' to signify the end of outputs.

样例输入

样例输出

20
60
*

简单模拟

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <cstdlib>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/sTACK:1024000000,1024000000")
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos[j](-1.0)
#define ei exp(1)
#define PI 3.1415926535
#define ios() ios[j]::sync_with_stdio(true)
#define INF 1044266558
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
ll a[105],n,s,t,w;
int main()
{
    while(scanf("%lld",&n))
    {
        if(n==0) break;
        memset(a,0,sizeof(a));
        for(ll i=0;i<n;i++)
        {
            scanf("%lld%lld%lld",&s,&t,&w);
            for(ll j=s;j<t;j++)
            {
                a[j]+=w;
            }
        }
        ll ans=0;
        for(ll i=1;i<=100;i++)
        {
            ans=max(ans,a[i]);
        }
        printf("%lld
",ans);
    }
    printf("*
");
    return 0;
}

F. Overlapping Rectangles

There are

Although the problem looks simple at the first glance, it might take a while to figure out how to do it correctly. Note that the shape of the union can be very complicated, and the intersected areas can be overlapped by more than two rectangles.

Note:

(1) The coordinates of these rectangles are given in integers. So you do not have to worry about the floating point round-off errors. However, these integers can be as large as

(2) To make the problem easier, you do not have to worry about the sum of the areas exceeding the long integer precision. That is, you can assume that the total area does not result in integer overflow.

Input Format

Several sets of rectangles configurations. The inputs are a list of integers. Within each set, the first integer (in a single line) represents the number of rectangles, n, which can be as large as

These configurations of rectangles occur repetitively in the input as the pattern described above. An integer

Output Format

For each set of the rectangles configurations appeared in the input, calculate the total area of the union of the rectangles. Again, these rectangles might overlap each other, and the intersecting areas of these rectangles can only be counted once. Output a single star '*' to signify the end of outputs.

样例输入

样例输出

7
3
*

求多矩形面积，可能存在重合  扫描线

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <cstdlib>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/sTACK:1024000000,1024000000")
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos[j](-1.0)
#define ei exp(1)
#define PI 3.1415926535
#define ios() ios[j]::sync_with_stdio(true)
#define INF 1044266558
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 3000;
int n;
struct Seg 
{
    double l,r,h;
    int d;
    Seg(){}
    Seg(double l,double r,double h,int d):l(l),r(r),h(h),d(d){}
    bool operator<(const Seg& rhs) const {return h<rhs.h;}
}a[N];
int cnt[N<<2];
double sum[N<<2],all[N];
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
void push_up(int l,int r,int rt) 
{
    if(cnt[rt]) sum[rt]=all[r+1]-all[l];
    else if(l==r) sum[rt]=0;
    else sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}

void update(int L,int R,int v,int l,int r,int rt) 
{
    if(L<=l && r<=R) {
        cnt[rt]+=v;
        push_up(l,r,rt);
        return;
    }
    int m = l + r >> 1;
    if(L<=m) update(L,R,v,lson);
    if(R>m) update(L,R,v,rson);
    push_up(l,r,rt);
}
int main()
{
    ios_base::sync_with_stdio(0);
    int kase = 0;
    while(scanf("%d",&n))
    {
        if(n==0) break;
        for(int i=1;i<=n;++i) 
        {
            double x1,y1,x2,y2;
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            a[i]=Seg(x1,x2,y1,1);
            a[i+n]=Seg(x1,x2,y2,-1);
            all[i]=x1;all[i+n]=x2;
        }
        n<<=1;
        sort(a+1, a+1+n);
        sort(all + 1, all+1+n);
        int m=unique(all+1,all+1+n)-all-1;
        memset(cnt,0,sizeof(cnt));
        memset(sum,0,sizeof(sum));
        double ans=0;
        for(int i=1;i<n;++i) 
        {
            int l=lower_bound(all+1, all+1+m, a[i].l)-all;
            int r=lower_bound(all+1, all+1+m, a[i].r)-all;
            if(l<r) update(l,r-1,a[i].d,1,m,1);
            ans+=sum[1]*(a[i+1].h-a[i].h);
        }
        printf("%.lf
",ans);
    }
    printf("*
");
    return 0;
}

L. The Heaviest Non-decreasing Subsequence Problem

Let $s1s_{1}s1, s2s_{2}s2, ........., sns_{n}sn Each integer is is associated with a weight by the following rules:$

(1) If is is negative, then its weight is

(2) If is is greater than or equal to $sis_{i}si is si−10000s_{i}-10000si−10000 . For example, if sis_{i}si is 101011010110101, then is is reset to 101101101 and its weight is 555.$

(3) Otherwise, its weight is

A non-decreasing subsequence of $si1s_{i1}si1, si2s_{i2}si2, ........., siks_{ik}sik, with i1<i2 ... <iki_{1}<i_{2} ... <i_{k}i1<i2 ... <ik, such that, for all 1≤j<k1 leq j<k1≤j<k, we have sij<sij+1s_{ij}<s_{ij+1}sij<sij+1.$

A heaviest non-decreasing subsequence of

Write a program that reads a sequence of integers, and outputs the weight of its

heaviest non-decreasing subsequence. For example, given the following sequence:

The heaviest non-decreasing subsequence of the sequence is

We guarantee that the length of the sequence does not exceed $2∗1052*10^{5}2∗105$

Input Format

A list of integers separated by blanks: $s1s_{1}s1, s2s_{2}s2,.........,sns_{n}sn$

Output Format

A positive integer that is the weight of the heaviest non-decreasing subsequence.

样例输入

80 75 73 93 73 73 10101 97 -1 -1 114 -1 10113 118

样例输出

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <cstdlib>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/sTACK:1024000000,1024000000")
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos[j](-1.0)
#define ei exp(1)
#define PI 3.1415926535
#define ios() ios[j]::sync_with_stdio(true)
#define INF 1044266558
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
int dp[1000006],a[1000006],x;
int main()
{
    int n=0;
    while(scanf("%d",&x)!=EOF)
    {
        if(x>=10000)
        {
            for(int i=0;i<5;i++)
                a[++n]=x-10000;
        }
        else if(x>=0) a[++n]=x;
    }
    memset(dp,62,sizeof(dp));
    for(int i=1;i<=n;i++)
        dp[upper_bound(dp,dp+1000000,a[i])-dp]=a[i];
    printf("%d
",lower_bound(dp,dp+1000000,INF)-dp);
    return 0;
}

M. Frequent Subsets Problem

Output Format

The number of

样例输入

15 0.4
1 8 14 4 13 2
3 7 11 6
10 8 4 2
9 3 12 7 15 2
8 3 2 4 5

样例输出

暴力枚举子集

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <cstdlib>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/sTACK:1024000000,1024000000")
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos[j](-1.0)
#define ei exp(1)
#define PI 3.1415926535
#define ios() ios[j]::sync_with_stdio(true)
#define INF 1044266558
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
int n,val[55],x,k;
double f;
char ch;
int main()
{
    scanf("%d%lf",&n,&f);
    k=0;
    while(~scanf("%d%c",&x,&ch))
    {
        val[k]+=(1<<(x-1));
        if(ch=='
') k++;
    }
    int ans=0,cnt;
    int pos=ceil(f*k);
    for(int i=1;i<(1<<n);i++)
    {
        cnt=0;
        for(int j=0;j<k;j++)
        {
            if((i&val[j])==i) printf("%d %d
",i,val[j]),cnt++;
        }
        ans+=cnt>=pos?1:0;
    }
    printf("%d
",ans);
    return 0;
}

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原文地址：https://www.cnblogs.com/shinianhuanniyijuhaojiubujian/p/7591655.html