POJ1659 Frogs' Neighborhood
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题意:给定一个图n(2 < n < 10)个结点相邻的结点个数,判断是否可图。
分析:
看到这数据范围,满满的暴搜感,那既然zrt把它放到了图论题,我们寻求一些其他的算法。
yzy神犇自己想出来了一半的正解,膜拜膜拜...
据说有个定理叫havel定理,判定过程如下:
1.对当前数列排序,使其成递减。
2.把从x(2)开始的x(1)个数字-1.
3.重复1, 2步,直到序列出现负数(不可图),或全为0(可图)。
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int T, n, ok, ans[15][15];struct X {int to, d;bool operator < (const X &rhs) const {return d > rhs.d;}}x[15];int main() {scanf("%d", &T);while(T--) {memset(ans, 0, sizeof ans);scanf("%d", &n);for(int i = 1; i <= n; i++) scanf("%d", &x[i].d), x[i].to = i;while(1) {sort(x+1, x+1+n);if(x[1].d <= 0) {ok = x[n].d == 0;break;}for(int i = 2; i <= x[1].d + 1; i++) x[i].d--, ans[x[1].to][x[i].to] = 1, ans[x[i].to][x[1].to] = 1;x[1].d = 0;}if(ok) {puts("YES");for(int i = 1; i <= n; i++, printf("\n"))for(int j = 1; j <= n; j++)printf("%d ", ans[i][j]);} else puts("NO");printf("\n");}}
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