ZOJ 3732 Graph Reconstruction

題目連結
Let there be a simple graph with $N$ vertices but we just know the degree of each vertex. Is it possible to reconstruct the graph only by these information?

A simple graph is an undirected graph that has no loops (edges connected at both ends to the same vertex) and no more than one edge between any two different vertices. The degree of a vertex is the number of edges that connect to it.

Input
There are multiple cases. Each case contains two lines. The first line contains one integer $N$ $(2\le N\le 100)$, the number of vertices in the graph. The second line conrains $N$ integers in which the ith item is the degree of ith vertex and each degree is between $0$ and $N-1$(inclusive).
Output
If the graph can be uniquely determined by the vertex degree information, output “UNIQUE” in the first line. Then output the graph.

If there are two or more different graphs can induce the same degree for all vertices, output “MULTIPLE” in the first line. Then output two different graphs in the following lines to proof.

If the vertex degree sequence cannot deduced any graph, just output “IMPOSSIBLE”.

The output format of graph is as follows:

$NE$
$u_1{u}_2\dots u_E$
$v_1{v}_2\dots v_E$
Where $N$ is the number of vertices and $E$ is the number of edges, and { u i , v i } \{u_i,v_i\} {ui​,vi​} is the $i_{th}$ edge the the graph. The order of edges and the order of vertices in the edge representation is not important since we would use special judge to verify your answer. The number of each vertex is labeled from $1$ to $N$. See sample output for more detail.
Sample Input

1
0
6
5 5 5 4 4 3
6
5 4 4 4 4 3
6
3 4 3 1 2 0

Sample Output

UNIQUE
1 0


UNIQUE
6 13
3 3 3 3 3 2 2 2 2 1 1 1 5
2 1 5 4 6 1 5 4 6 5 4 6 4
MULTIPLE
6 12
1 1 1 1 1 5 5 5 6 6 2 2
5 4 3 2 6 4 3 2 4 3 4 3
6 12
1 1 1 1 1 5 5 5 6 6 3 3
5 4 3 2 6 4 3 2 4 2 4 2
IMPOSSIBLE

Havel-Hakimi 定理：由非負整陣列成的不遞增序列 $s:d_1,d_2,\dots d_n(n\ge 2,d_1\ge 1)$ 是可圖的，當且僅當序列
$$s_1:d_2-1,d_3-1,\cdots ,d_{d_1+1}-1,d_{d_1+2},\cdots ,d_n$$
是可圖的。序列 $s_1$ 中有 $n-1$ 個非負整數， $s$ 序列中 $d_1$後的前 $d_1$個度數（即 $d_2\sim d_{d_1+1}$）減 $1$ 後構成 $s_1$ 中的前 $d_1$ 個數。對應到構造解的過程即為將節點 $1$ 向節點 $2\sim d_1+1$ 各連一條邊。如果$d_{d_1+1}=d_{d_1+2}$，那麼存在另一種建圖方案，其中$1$連$d_1+2$而不是$d_1+1$，顯然兩種方案一定不相同。
至於不存在$d_{d_1+1}=d_{d_1+2}$那麼解唯一，emm…不會證

#include<bits/stdc++.h>

#define pii pair<int,int>
#define ce(i, r) i==r?'\n':' '
#define fi first
#define se second
using namespace std;

inline int qr() {
    int f = 0, fu = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-')fu = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        f = (f << 3) + (f << 1) + c - 48;
        c = getchar();
    }
    return f * fu;
}

const int N = 105, M = N * N;
pii p[2][N], e[2][M];
int n, t[2];

int main() {
    while (~scanf("%d", &n)) {
        t[0] = t[1] = 0;
        for (int i = 1; i <= n; i++)p[0][i] = {qr(), i};
        bool suc1 = true, suc2 = false;
        int m = n;
        sort(p[0] + 1, p[0] + 1 + m, greater<pii >());
        memcpy(p[1], p[0], sizeof(pii) * (n + 1));
        for (int i = 1; i <= m && p[0][i].fi; i++) {
            while (!p[0][m].fi)m--;
            if (i + p[0][i].fi > m) {
                suc1 = false;
                break;
            }
            if (i + p[0][i].fi < m && p[1][i + p[0][i].fi].fi == p[1][i + p[0][i].fi + 1].fi)
                swap(p[1][i + p[0][i].fi], p[1][i + p[0][i].fi + 1]), suc2 = true;
            for (int j = i + 1; j <= i + p[0][i].fi; j++)
                p[0][j].fi--, e[0][++t[0]] = {p[0][i].se, p[0][j].se};
            for (int j = i + 1; j <= i + p[1][i].fi; j++)
                p[1][j].fi--, e[1][++t[1]] = {p[1][i].se, p[1][j].se};
            sort(p[0] + i + 1, p[0] + 1 + m, greater<pii >());
            sort(p[1] + i + 1, p[1] + 1 + m, greater<pii >());
        }
        if (!suc1)puts("IMPOSSIBLE");
        else if (!suc2) {
            puts("UNIQUE");
            printf("%d %d\n", n, t[0]);
            for (int i = 1; i <= t[0]; i++)printf("%d%c", e[0][i].fi, ce(i, t[0]));
            for (int i = 1; i <= t[0]; i++)printf("%d%c", e[0][i].se, ce(i, t[0]));
            if (!t[0])puts("\n");
        } else {
            puts("MULTIPLE");
            printf("%d %d\n", n, t[0]);
            for (int i = 1; i <= t[0]; i++)printf("%d%c", e[0][i].fi, ce(i, t[0]));
            for (int i = 1; i <= t[0]; i++)printf("%d%c", e[0][i].se, ce(i, t[0]));
            printf("%d %d\n", n, t[1]);
            for (int i = 1; i <= t[1]; i++)printf("%d%c", e[1][i].fi, ce(i, t[1]));
            for (int i = 1; i <= t[1]; i++)printf("%d%c", e[1][i].se, ce(i, t[1]));
        }
    }
    return 0;
}