用到蛋疼的sort

Wilbur and Points

 
Codeforces Round #331 (Div. 2)

C. Wilbur and Points

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Wilbur is playing with a set of n points on the coordinate plane. All points have non-negative integer coordinates. Moreover, if some point (x,
y) belongs to the set, then all points (x',
y'), such that 0 ≤ x' ≤ x and
0 ≤ y' ≤ y also belong to this set.

Now Wilbur wants to number the points in the set he has, that is assign them distinct integer numbers from
1 to n. In order to make the numbering
aesthetically pleasing, Wilbur imposes the condition that if some point (x,
y) gets number i, then all (x',y') from the set, such that
x' ≥ x and
y' ≥ y must be assigned a number not less than
i. For example, for a set of four points (0,
0), (0,
1), (1, 0) and (1,
1), there are two aesthetically pleasing numberings. One is
1, 2,
3, 4 and another one is
1, 3, 2,
4.

Wilbur's friend comes along and challenges Wilbur. For any point he defines it's
special value as s(x, y) = y - x. Now he gives Wilbur some
w1,
w2,..., wn, and asks him to find an aesthetically pleasing numbering of the points in the set, such that the point that gets number
i has it's special value equal to
wi, that is
s(xi, yi) = yi - xi = wi.

Now Wilbur asks you to help him with this challenge.

Input
The first line of the input consists of a single integer
n (1 ≤ n ≤ 100 000) — the number of points in the set Wilbur is playing with.

Next follow n lines with points descriptions. Each line contains two integers
x and y (0 ≤ x, y ≤ 100 000), that give one point in Wilbur's set. It's guaranteed that all points are distinct. Also, it is guaranteed
that if some point (x,
y) is present in the input, then all points (x',
y'), such that 0 ≤ x' ≤ x and
0 ≤ y' ≤ y, are also present in the input.

The last line of the input contains n integers. The
i-th of them is wi ( - 100 000 ≤ wi ≤ 100 000) — the required
special value of the point that gets number i in any aesthetically pleasing numbering.

Output
If there exists an aesthetically pleasant numbering of points in the set, such that
s(xi, yi) = yi - xi = wi,
then print "YES" on the first line of the output. Otherwise, print "NO".

If a solution exists, proceed output with n lines. On the
i-th of these lines print the point of the set that gets number
i. If there are multiple solutions, print any of them.

Sample test(s)

Input

5
2 0
0 0
1 0
1 1
0 1
0 -1 -2 1 0

Output

YES
0 0
1 0
2 0
0 1
1 1

Input

3
1 0
0 0
2 0
0 1 2

Output

NO

代码：  

    #include <cstdio>

    #include <cstring>

    #include <cmath>

    #include <cstdlib>

    #include <algorithm>

    #include <queue>

    #include <stack>

    #include <map>

    #include <vector>

    #define INF 0x3f3f3f3f

    #define eps 1e-4

    #define MAXN (200000+10)

    #define MAXM (1000000)

    #define Ri(a) scanf("%d", &a)

    #define Rl(a) scanf("%lld", &a)

    #define Rf(a) scanf("%lf", &a)

    #define Rs(a) scanf("%s", a)

    #define Pi(a) printf("%d\n", (a))

    #define Pf(a) printf("%lf\n", (a))

    #define Pl(a) printf("%lld\n", (a))

    #define Ps(a) printf("%s\n", (a))

    #define W(a) while(a--)

    #define CLR(a, b) memset(a, (b), sizeof(a))

    #define MOD 100000007

    #define LL long long

    #define lson o<<1, l, mid

    #define rson o<<1|1, mid+1, r

    #define ll o<<1

    #define rr o<<1|1

    using namespace std;

    struct Node{

        int x, y, val, id, rec;

    };

    Node num1[MAXN], num2[MAXN];

    bool cmp(Node a, Node b)

    {

        if(a.val != b.val)

            return a.val < b.val;

        else if(a.x != b.x)

            return a.x < b.x;

        else

            return a.y < b.y;

    }

    bool cmp1(Node a, Node b)

    {

        if(a.val != b.val)

            return a.val < b.val;

        else

            return a.id < b.id;

    }

    bool cmp2(Node a, Node b){

        return a.id < b.id;

    }

    int main()

    {

        int n; Ri(n);

        for(int i = 0; i < n; i++)

        {

            Ri(num1[i].x); Ri(num1[i].y);

            num1[i].val = num1[i].y - num1[i].x;

            num1[i].id = i;

        }

        sort(num1, num1+n, cmp);

        for(int i = 0; i < n; i++)

        {

            Ri(num2[i].val);

            num2[i].id = i;

        }

        sort(num2, num2+n, cmp1);

        bool flag = true;

        for(int i = 0; i < n; i++)

        {

            if(num1[i].val != num2[i].val)

            {

                flag = false;

                break;

            }

            else

                num2[i].rec = num1[i].id;

        }

        if(flag)

        {

            sort(num2, num2+n, cmp2);

            sort(num1, num1+n, cmp2);

            for(int i = 1; i < n; i++)

            {

                if(!(num1[num2[i].rec].x > num1[num2[i-1].rec].x || num1[num2[i].rec].y > num1[num2[i-1].rec].y))

                {

                    flag = false;

                    break;

                }

            }

            if(flag)

            {

                printf("YES\n");

                for(int i = 0; i < n; i++)

                    printf("%d %d\n", num1[num2[i].rec].x, num1[num2[i].rec].y);

            }

            else

                printf("NO\n");

        }

        else

            printf("NO\n");

        return 0;

    }