P1963 [NOI2009]变换序列 倒叙跑匈牙利算法

题意

构造一个字典序最小的序列T，使得 $Dis（i， T_i） = d_i$，其中i是从0开始的，$Dis(x,y)=min{∣x−y∣,N−∣x−y∣} $，$d_i$由题目给定。

思路

二分图匹配，把左边的看成i，右边看成$T_i$，对于固定的i和d，Ti是由两种可能的，连上有向边即可。
至于字典序要最小，怎么做呢，我们可以反着跑匈牙利算法，就是从n-1跑到0，这样小一点的i，可以直接拿走大一点的i刚匹配的较小的值。

 

#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>

using namespace std;
#define lson (l, mid, rt << 1)
#define rson (mid + 1, r, rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<int, pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define boost                    \
    ios::sync_with_stdio(false); \
    cin.tie(0)
#define rep(a, b, c) for (int a = (b); a <= (c); ++a)
#define max3(a, b, c) max(max(a, b), c);
#define min3(a, b, c) min(min(a, b), c);

const ll oo = 1ll << 17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9 + 7;
const double esp = 1e-8;
const double PI = acos(-1.0);
const double PHI = 0.61803399; //黄金分割点
const double tPHI = 0.38196601;

template <typename T>
inline T read(T &x) {
    x = 0;
    int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x = f ? -x : x;
}
inline void cmax(int &x, int y) {
    if (x < y) x = y;
}
inline void cmax(ll &x, ll y) {
    if (x < y) x = y;
}
inline void cmin(int &x, int y) {
    if (x > y) x = y;
}
inline void cmin(ll &x, ll y) {
    if (x > y) x = y;
}

/*-----------------------showtime----------------------*/

const int maxn = 1e4 + 9;
struct E {
    int v, nxt;
} edge[4 * maxn];
int head[maxn], gtot;
void addedge(int u, int v) {
    edge[gtot].v = v;
    edge[gtot].nxt = head[u];
    head[u] = gtot++;
}
int vis[maxn], pt[maxn], py[maxn];
bool dfs(int u) {
    for (int i = head[u]; ~i; i = edge[i].nxt) {
        int v = edge[i].v;
        if (vis[v] == 0) {
            vis[v] = 1;
            if (pt[v] == 0 || dfs(pt[v])) {
                pt[v] = u;
                py[u] = v;
                return true;
            }
        }
    }
    return false;
}
int main() {
    int n;
    scanf("%d", &n);
    memset(head, -1, sizeof(head));
    priority_queue<int> que;
    while (!que.empty()) que.pop();
    for (int i = 0; i < n; i++) {
        int d;
        scanf("%d", &d);
        int a = i - d;

        if (a >= 0 && a < n) que.push(a);
        a = i + d;
        if (a >= 0 && a < n) que.push(a);
        int b = i - (n - d);
        if (b >= 0 && b < n) que.push(b);
        b = i + (n - d);
        if (b >= 0 && b < n) que.push(b);
        while (!que.empty()) {
            int t = que.top();
            que.pop();
            addedge(i, t);
        }
    }
    int res = n;
    for (int i = n - 1; i >= 0; i--) {
        memset(vis, 0, sizeof(vis));
        if (dfs(i))
            res = i;
        else
            break;
    }
    if (res == 0) {
        for (int i = 0; i < n; i++) printf("%d ", py[i]);
        puts("");
    } else
        puts("No Answer");
    return 0;
}