Codeforces 1009G Allowed Letters 最大流转最小割 sosdp

Posted 2021-02-16 cjlhy

Allowed Letters

最直观的想法是贪心取， 然后网络流取check可不可行， 然后T了。

想到最大流可以等于最小割， 那么我们状压枚举字符代表的6个点连向汇点是否断掉，

然后再枚举64个本质不同的位置， 是否需要切段原点联想它的边， 复杂度电磁check复杂度64 * 64

用sosdp能优化到64 * 6

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()

using namespace std;

const int N = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

int n, m, mask[N], c[N], cnt[N], sos[N];
char s[N], t[N], ans[N];

bool check() {
    int sum = 0, ret = inf;
    for(int i = 0; i < 6; i++) sum += c[i];
    for(int i = 0; i < 64; i++) sos[i] = cnt[i];
    for(int i = 0; i < 6; i++)
        for(int j = 0; j < 64; j++)
            if(j >> i & 1) sos[j] += sos[j ^ (1 << i)];
    for(int s1 = 0; s1 < 64; s1++) {
        int tmp = 0;
        for(int i = 0; i < 6; i++)
            if(s1 >> i & 1) tmp += c[i];
        tmp += sum - sos[s1];
        chkmin(ret, tmp);
    }
    return sum == ret;
}

inline int getId(char c) {
    return c - ‘a‘;
}

int main() {
    scanf("%s", s + 1);
    n = strlen(s + 1);
    for(int i = 1; i <= n; i++) c[getId(s[i])]++;
    scanf("%d", &m);
    while(m--) {
        int p; scanf("%d%s", &p, t + 1);
        for(int i = 1; t[i]; i++) mask[p] |= 1 << getId(t[i]);
    }
    for(int i = 1; i <= n; i++) {
        if(!mask[i]) mask[i] = 63;
        cnt[mask[i]]++;
    }
    for(int i = 1; i <= n; i++) {
        bool flag = false;
        for(int j = 0; j < 6; j++) {
            if(c[j] && mask[i] >> j & 1) {
                c[j]--;
                cnt[mask[i]]--;
                flag = check();
                if(flag) {
                    ans[i] = ‘a‘ + j;
                    break;
                }
                c[j]++;
                cnt[mask[i]]++;
            }
        }
        if(!flag) return puts("Impossible"), 0;
    }
    ans[n + 1] = ‘‘;
    puts(ans + 1);
    return 0;
}

/*
*/