URAL 1105 Observers Coloring （构造法）

aozil_yang

于 2016-10-31 21:05:21 发布

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分类专栏： URAL 文章标签： URAL 构造法 C++

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解决一个数学竞赛问题，通过构造法确定考场内特定时间段仅有单一染色监考员的最长时间，确保该时间不短于考生答题时间的三分之二。

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大体题意：

你在一个考场里做题，做题时间是t0~t1，有n (n <= 10000)个监考员，告诉你每一个监考员的入场时间和出场时间，现在你可以给一定量的监考员进行染色，使得考场内只有一个且染色监考员的总时间不小于你做题时间的2/3，如果可行输出你染色的监考员，如果不可行输出0！输入保证在你做题时间至少有一个监考员！

思路：

构造法！

知道了每个监考员的入场时间和出场时间，我们可以把它看成一个区间！最后要求你保留一些区间，使得这个区间只有一个线段的数值总和不小于考试时间的2/3.

我们把去掉后没有空白的区间给删掉！这样最后给剩下的区间按照左端点进行排序，那么第i 个区间和第i+1个区间一定相交，并且第i 个区间一定不和第i+2个区间相交，否则第i+2个区间或者第i+1个区间可以被删掉！

这样我们选出所有奇数号的区间，这些区间一定都不相交，加起来如果大于2/3 输出！

如果上述情况不合适那么我们在加偶数号区间，这些区间同样也都不想交，加起来如果大于2/3 输出！

如果还是不大于2/3

那么我们来看看全选的时间!

令a = 奇数号的区间时间总和！

令b = 偶数号区间的时间总和！

令T 为你的做题时间！

如果全选，那么剩下的“单独时间” 的总和为 T-a + T - b（这个式子画个图就很明了了！）

现在a < 2*T/3 b < 2*T/3

那么单独时间为2*T - (a+b) >= 2*T/3

所以说如果上述奇偶都不合理的情况下全选一定有解！所以此题一定有解！

因为是浮点数注意好精度控制就好了!

贴个图帮助大家理解：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;
const double eps = 1e-10;
const double inf = 1e18;
const int maxn = 10000 + 7;
double t0,t1;
double dcmp(double a,double b){ // 比较浮点数大小
    if (fabs(a-b) < eps) return 0;
    if (a > b) return 1;
    return -1;
}
struct Node{
    double l,r;
    int id;
    void read() { scanf("%lf %lf",&l, &r);}
    bool operator < (const Node& rhs) const {
        return dcmp(l,rhs.l) == -1 || (dcmp(l,rhs.l) == 0 && dcmp(r,rhs.r) == 1);
    }
}p[maxn];
vector<int>ans;
int main(){
    while(~scanf("%lf %lf",&t0, &t1)){
        int n;
        ans.clear();
        double T = 2.0*(t1-t0)/3.0; // 总时间值！
        scanf("%d",&n);
        for (int i = 0; i < n; ++i){
            p[i].read();
            p[i].id = i+1;
        }
        sort(p,p+n);
        int len = 0;
        for (int i = 0; i < n; ++i){
            bool ok = 0;
            if (len == 0){
                ans.push_back(i);
                ok = 1;
                len++;
            }
            else if (len == 1){
                int id = ans[0];
                if (dcmp(p[i].l,p[id].l) == 1 && dcmp(p[i].r,p[id].r) == 1) ans.push_back(i),ok = 1,++len;
            }
            else {
                int laa = ans[len-2];
                int la = ans[len-1];
                if (dcmp(p[laa].r,p[i].l) == -1 && dcmp(p[i].r,p[la].r) == 1){
                    ans.push_back(i);
                    ++len;
                    ok = 1;
                }
                else if (dcmp(p[laa].r,p[i].l) != -1 && dcmp(p[i].r,p[la].r) != -1){
                    ans.pop_back();
                    ans.push_back(i);
                    ok = 1;
                }

            }
            if (!ok) continue;
            int cur = i;
            while(i+1 < n && dcmp(p[i+1].l,p[cur].l) == 0)++i;
        }
        double sum1 = 0,sum2 = 0; //  分奇数偶数的形式来讨论！
        int cnt1=0,cnt2=0;
        for (int i = 0; i < ans.size(); i+=2){
            int id = ans[i];
            sum1 += p[id].r - p[id].l;
            ++cnt1;
        }
        if (dcmp(sum1,T) != -1) {
            printf("%d\n",cnt1);
            for (int i = 0; i < ans.size(); i+=2){
                int id = ans[i];
                printf("%d\n",p[id].id);
            }
        }
        else {
            for (int i = 1; i < ans.size(); i+=2){
                int id = ans[i];
                sum2 += p[id].r - p[id].l;
                ++cnt2;
            }
            if (dcmp(sum2,T) != -1) {
                printf("%d\n",cnt2);
                for (int i = 1; i < ans.size(); i+=2){
                    int id = ans[i];
                    printf("%d\n",p[id].id);
                }
            }
            else {
                printf("%d\n",ans.size());
                for (int i = 0; i < ans.size(); ++i){
                    printf("%d\n",p[ans[i]].id);
                }
            }
        }
    }
    return 0;
}
/**
0 6
3
0 3
2 5 4 6
**/

1105. Observers Coloring

Time limit: 0.5 second
Memory limit: 64 MB

Nikifor told us that once he solved problem at mathematical tournament of S.Petersburg's secondary school N239 in 1994. Nikifor said that he solved a problem from the moment T 0 to the moment T 1. He remembers that N observers appeared in the room. The i-th observer entered the room at the moment t 0, i and went out at the moment t 1, i. At every moment there was at least one observer in the room.

When the tournament was finished, Nikifor claimed that it is possible to color some observers, and the summary time when there was only one colored observer in the room is not less than 2/3 of the time when Nikifor solved problem.

You are to answer whether Nikifor right or not.

Input

The first line of input contains real numbers T 0 and T 1 ( T 0 < T 1). The second line contains number N — number of observers ( N < 10000). Next N lines contain real numbers t 0, i and t 1, i ( T 0 ≤ t 0, i < t 1, i ≤ T 1).

Output

If Nikifor is not right output should contain the only number 0. If Nikifor is right you should write to the first line the quantity of colored observers, and next lines should contain their numbers. Do not write more than one number in a line. You may write these numbers in any order. If there are more than one solution exist you may find any of them.