首页 > 代码库 > UVALive 6088 Approximate Sorting 构造题

UVALive 6088 Approximate Sorting 构造题

题目链接:点击打开链接

题意:

给定一个n*n的01矩阵

我们跑一下样例==

4
0111
0000
0100
0110

  0123

\|____
0|0111
1|0000
2|0100
3|0110

用0-n-1随便构造一个序列:

如:1230

我们计算1230的权值 :

int ans = 0;

对于个位0,我们查找第0行:0111,0前面的数有123, 则ans += a[0][1], ans+=a[0][2], ans+=a[0][3]

对于十位3,我们查找第3行:0110,3前面的数有12, ans+=a[3][1], ans+=a[3][2]

如此得到ans = 6

问:

构造一个这样的序列使得ans最小,若有多个这样的序列则输出序列字典序最小的。

#include <cstdio>
#include <iostream>
#include <string>
#include <map>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int Inf = 1e9;
typedef long long ll;
const int N = 18;
const int S = 1 << N;
int d[S], one[S];
int w[N], n, mx;
int out[N], dep;
map<int, int> id;

int dp(int s) {
	if (~d[s])
		return d[s];
	d[s] = Inf;
	int i, to, x = s;
	while (x>0) {
		to = (x&(x-1)) ^ x;
		i = id[to];
		d[s] = min(d[s], dp(s^(1<<i)) + one[w[i]&s]);
		x = x^to;
	}
	return d[s];
}
void dfs(int dep, int cur, int g) {
	if (dep == N) {
		one[cur] = g;
	} else {
		dfs(dep+1, cur*2+1, g+1);
		dfs(dep+1, cur*2, g);
	}
}
void path(int s) {
	if (s != 0) {
		for (int i = 0; i < n; ++i)
			if (s >> i & 1) {
				if (dp(s^(1 << i)) + one[w[i]&s] == dp(s) ) {
					out[dep++] = i;
					path(s^(1 <<i));
					return ;
				}
			}
	}
}
char s[N];
void work() {
	memset(w, 0, sizeof w);
	for (int i = 0; i < n; ++i) {
		scanf("%s", s);
		for (int j = 0; j < n; ++j)
			if (s[j] == '1' && j != i)
				w[j] |= 1 << i;
	}
	mx = (1 << n) - 1;
	memset(d, -1, sizeof d);
	d[0] = 0;
	int ans = dp(mx);
	dep = 0;
	path(mx);
	for (int i = 0; i < dep; ++i) {
		if (i > 0)
			putchar(' ');
		printf("%d", out[i]);
	}
	puts("");
	printf("%d\n", ans);
}
int main() {
	dfs(0, 0, 0);
	for (int i = 0; i < N; ++i)
		id[1 << i] = i;
	while (~scanf("%d", &n)) {
		if (0 == n)
			break;
		work();
	}
	return 0;
}


UVALive 6088 Approximate Sorting 构造题