POJ1659 Frogs' Neighborhood(Havel–Hakimi定理)
题意
\(T\)组数据,给出\(n\)个点的度数,问是否可以构造出一个简单图
Sol
Havel–Hakimi定理:
- 给定一串有限多个非负整数组成的序列,是否存在一个简单图使得其度数列恰为这个序列。
令\(S=(d_1,d_2,\dots,d_n)\)为有限多个非负整数组成的非递增序列。 S可简单图化当且仅当有穷序列\(S’=(d_2-1,d_3-1,...,d(d_1+1)-1,d(d_1+2),...,d_n)\)只含有非负整数且是可简单图化的。
最后判断一下是否都是零就好了
感觉这个算法。。就是个贪心吧。。
当然判断这类问题的可行性还有另外一种方法:Erdős–Gallai定理
令\(S=(d_1,d_2,...,d_n)\)为有限多个非负整数组成的非递增序列。\(S\)可简单图化当且仅当这些数字的和为偶数,并且
\(\sum_{i = 1}^k d_i \leqslant k(k - 1) + \sum_{i = k + 1}^n min(d_i, k)\)
对所有\(1 \leqslant k \leqslant n\)都成立
不过这个好像没办法输出方案??。。。
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
using namespace std;
const int MAXN = 1e5 + 10, INF = 1e9 + 7;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int T, N, reach[101][101], sum = 0;
Pair a[MAXN];
void init() {
memset(reach, 0, sizeof(reach));
sum = 0;
}
int main() {
// freopen("a.in", "r", stdin);
T = read();
while(T--) {
init();
N = read();
for(int i = 1; i <= N; i++) a[i] = MP(read(), i), sum += a[i].fi;
if(sum % 2 != 0) {puts("NO\n"); continue;}
bool f = 0;
for(int i = 1; i <= N; i++) {
sort(a + i, a + N + 1, greater<Pair>());
if(a[i].fi <= 0) continue;
for(int j = i + 1; j <= i + a[i].fi; j++) a[j].fi -= 1, reach[a[i].se][a[j].se] = 1, reach[a[j].se][a[i].se] = 1;
a[i].fi = 0;
}
for(int i = 1; i <= N; i++) if(a[i].fi != 0) {puts("NO\n"); f = 1; break;}
if(f) continue;
puts("YES");
for(int i = 1; i <= N; i++, puts(""))
for(int j = 1; j <= N; j++)
printf("%d ", reach[i][j]);
puts("");
}
}
/*
1
6
4 3 1 4 2 0
*/
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