Codeforces 500B - New Year Permutation (思维)

题意

要得到尽量从小到大排列的序列,但是只有mp[i][j] = 1的时候才能交换两个位置。问能得到的最小序列。

思路

显然应该把小的尽量往前换,我是先对图跑一遍Floyd,然后对每个位置从小到大扫描一遍,能换就换。hcbbt巨巨用了并查集,时间比我少一半Orz。

代码

#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <iomanip>
#include <cmath>
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
#define BitCountll(x) __builtin_popcountll(x)
#define LeftPos(x) 32 - __builtin_clz(x) - 1
#define LeftPosll(x) 64 - __builtin_clzll(x) - 1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const double eps = 1e-8;
const int MAXN = 300 + 10;
const int MOD = 1000007;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
 
int mp[MAXN][MAXN];
int pos[MAXN], val[MAXN], n;
 
void Floyd()
{
    for (int k = 1; k <= n; k++)
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                if (mp[i][j] == 0) mp[i][j] = mp[i][k] & mp[k][j];
}
 
int main()
{
    //ROP;
    int i, j;
    scanf("%d", &n);
    for (i = 1; i <= n; i++)
    {
        int tmp;
        scanf("%d%*c", &tmp);
        pos[tmp] = i, val[i] = tmp;
    }
    for (i = 1; i <= n; i++)
    {
        for (j = 1; j <= n; j++)
        {
            char tmp;
            scanf("%c", &tmp);
            mp[i][j] = tmp - '0';
        }
        getchar();
    }
    Floyd();
    int conti = 1;
    for (i = 1; i <= n; i++)
        for (j = conti; j <= n; j++)
        {
            if (mp[i][pos[j]])
            {
                int oriPos = pos[j];
                val[oriPos] = val[i];
                pos[val[i]] = oriPos;
                val[i] = j;
                pos[j] = i;
 
            }
        }
    for (i = 1; i <= n; i++)
    {
        if (i == 1) printf("%d", val[i]);
        else printf(" %d", val[i]);
    }
    puts("");
    return 0;
}

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