【赛后补题】ccpc2107秦皇岛H题

ZOJ上可以补

赛场上偷听到了隔壁的匈牙利

然后一波操作，未曾想匈牙利之后操作如此复杂

毕竟金牌题，技不如人，甘拜下风

思路： 奇数和偶数分别算作一个集合，二分图构造完毕

烦人的就是那个1，1+1是质数，1+2也是质数，先1内部考虑，然后1+2

（写这段话的时候发现我想错了）

NONONO应该先考虑1+2，再考虑1+1

zoj数据没那么严，不想改了，各位随意

———-吉老师纠正

1+2的在二分图中已经判断了，太好了，我又写了一堆冗余代码，

wtf（）

所以剩下的就是，很多很多个1和一堆单身狗们（该说法来自匈牙利算法趣写）

#include <bits/stdc++.h>using namespace std;typedef long long LL;const int N = 3e3 + 7;//-------------------------prime--------------------------------const int PRIME_N = 2e6 + 7;int prime[PRIME_N], primesize;bool isprime[PRIME_N + 7];void getprime(int listsize){    memset(isprime, 1, sizeof(isprime));    isprime[1] = false;    primesize = 0;    for(int i=2; i < listsize; i++){        if(isprime[i]) prime[++primesize]=i;        for(int j = 1; j <= primesize && i*prime[j] < listsize; j++){            isprime[i*prime[j]] = false;            if(i%prime[j] == 0) break;        }    }}//-------------------------hungary-------------------------------bool marry[N], adore[N];int arr[N], boy[N], gir[N], B, G;struct Hungary{    vector <int> G[N];    bool used[N];// main里面记得memset    int girl[N], n;    inline void init(int _n){        n = _n;        for (int i = 0; i <= n; i++) G[i].clear();    }    inline void addEdge(const int &u, const int &v){        G[u].push_back(v);    }    bool Find(int x){        for (int i = 0; i < G[x].size(); i++){    //扫描每个妹子            int j = G[x][i];            if (used[j]) continue;            // 如果有暧昧并且还没有标记过            // (这里标记的意思是这次查找曾试图改变过该妹子的归属问题，            // 是没有成功，所以就不用瞎费工夫了）            used[j] = 1;            if (girl[j] == 0 || Find(girl[j])) {                //名花无主或者能腾出个位置来，这里使用递归                girl[j] = x;                marry[boy[x]] = marry[gir[j]] = true; // this problem                return true;            }        }        return false;    }    inline int hungary(const int &n){        int all = 0;        memset(girl, 0, sizeof girl);        memset(marry, false, sizeof marry); // this problem        for (int i = 1; i <= n; i++) {            memset(used, 0, sizeof(used)); //这个在每一步中清空            if (Find(i)) all += 1;        }        return all;    }} hg;inline int wtf(const int &n, const int &B, const int &k){    int mat = hg.hungary(B);    //printf("mat = %d\n", mat);    if (mat >= k) return k * 2;    int cnt1 = 0, idx = n; // number of one    for (int i = n; i >= 1; i--){        if (arr[i] != 1) {idx = i; break;}        if ((!marry[i])) cnt1++;        marry[i] = true;    }    //printf("cnt1 = %d\n", cnt1);    mat += cnt1 >> 1; // 剩下的1    if (mat >= k) return k*2;    int left = cnt1 & 1; // 单身狗们    for (int i = 1; i <= idx; i++){        left += (!marry[i]) & adore[i];    }    if (mat + left <= k) return mat*2+left;    return mat*2 + (k - mat);}int main(){    //freopen("in.txt", "r", stdin);    getprime(2e6 + 7);    int _;    scanf("%d", &_);    for (int n, k; _--;) {        scanf("%d%d", &n, &k);        for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);        sort(arr + 1, arr + n+1, greater<int>());        B = G = 0;        for (int i = 1; i <= n; i++){            if (arr[i] & 1) boy[++B] = i;            else gir[++G] = i;        }        hg.init(B);        memset(adore, false, sizeof adore);        for (int i = 1; i <= B; i++){            for (int j = 1; j <= G; j++){                int num = arr[boy[i]] + arr[gir[j]];                if (isprime[num]) {                    hg.addEdge(i, j);                    adore[boy[i]] = adore[gir[j]] = true;                }            }        }        int ans = wtf(n, B, k);        printf("%d\n", ans);    }    return 0;}

吉老师好帅，，，

网络流跑二分图会快很多,

150ms

#include <bits/stdc++.h>using namespace std;typedef long long LL;const int N = 3e3 + 7;const int INF=0x3f3f3f3f;//-------------------------prime--------------------------------const int PRIME_N = 2e6 + 7;int prime[PRIME_N], primesize;bool isprime[PRIME_N + 7];void getprime(int sz){    memset(isprime, 1, sizeof(isprime));    isprime[1] = false;    primesize = 0;    for (int i = 2; i < sz; i++){        if (isprime[i]) prime[++primesize]=i;        for (int j = 1; j <= primesize && i*prime[j] < sz; j++){            isprime[i*prime[j]] = false;            if (i % prime[j] == 0) break;        }    }}bool marry[N], adore[N];int arr[N], boy[N], gir[N], B, G;//------------------------Dinic-maxFlow--------------------------struct gragh{    struct Edge{        int from, to, cap, flow;        Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){}    };    int n, s, t;    vector <Edge> edges;    vector < int> G[N]; //gragh    bool vis[N]; //use when bfs    int d[N],cur[N];//dist,now edge,use in dfs    inline void init(int _n, int _s, int _t){        n = _n; s = _s, t = _t; edges.clear();        for (int i = 0; i <= n; i++) G[i].clear();    }    inline void addEdge(int from, int to, int cap){        edges.push_back(Edge(from,to,cap,0));        edges.push_back(Edge(to,from, 0 ,0));        int top = edges.size();        G[from].push_back(top-2);        G[ to ].push_back(top-1);    }    inline bool BFS(){        memset(vis, 0, sizeof(vis));        queue <int> Q;        d[s]=0;vis[s]=1;        for (Q.push(s); !Q.empty();){            int x = Q.front(); Q.pop();            for (int i = 0; i < G[x].size(); i++){                Edge &e = edges[G[x][i]];                if (vis[e.to] || e.cap <= e.flow)continue;                vis[e.to] = 1;                d[e.to] = d[x] + 1;                Q.push(e.to);            }        }        return vis[t];    }    inline int DFS(const int& x,int a){        //printf("dfs:%d,%d\n",x,a);        if (x==t||a==0){return a;}        int flow = 0, f;        for (int& i = cur[x]; i < G[x].size(); i++){            Edge& e = edges[G[x][i]];            if (d[x] + 1 != d[e.to]) continue;            if ((f=DFS(e.to,min(a,e.cap-e.flow)))<=0)continue;            e.flow += f;            edges[G[x][i]^1].flow -= f;//反向边            flow+=f; a-=f;            marry[x] = marry[e.to] = true;            if (a==0) break;        }        return flow;    }    inline int maxFlow(){return maxFlow(s,t);}    inline int maxFlow(const int& s, const int& t){        //if (edges.size() > N*N>>1) return 233;        int flow = 0;        while (BFS()){            memset(cur, 0, sizeof(cur));            int f = DFS(s, INF);            flow += f ;        }        return flow;    }} g;inline int wtf(const int &n, const int &B, const int &k){    memset(marry, false, sizeof marry);    int mat = g.maxFlow();    //printf("mat = %d\n", mat);    if (mat >= k) return k * 2;    int cnt1 = 0, idx = n; // number of one    for (int i = n; i >= 1; i--){        if (arr[i] != 1) {idx = i; break;}        if ((!marry[i])) cnt1++;        marry[i] = true;    }    //printf("cnt1 = %d\n", cnt1);    mat += cnt1 >> 1; // 剩下的1    if (mat >= k) return k*2;    int left = cnt1 & 1; // 单身狗们    for (int i = 1; i <= idx; i++){        left += (!marry[i]) & adore[i];    }    if (mat + left <= k) return mat*2+left;    return mat*2 + (k - mat);}int main(){    //freopen("in.txt", "r", stdin);    getprime(2e6 + 7);    int _;    scanf("%d", &_);    for (int n, k; _--;) {        scanf("%d%d", &n, &k);        for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);        sort(arr + 1, arr + n + 1, greater<int>());        B = G = 0;        g.init(n + 1, 0, n + 1);        for (int i = 1; i <= n; i++){            if (arr[i] & 1) {                boy[++B] = i;                g.addEdge(0, i, 1);            } else {                gir[++G] = i;                g.addEdge(i, n+1, 1);            }        }        memset(adore, false, sizeof adore);        for (int i = 1; i <= B; i++){            for (int j = 1; j <= G; j++){                int num = arr[boy[i]] + arr[gir[j]];                if (isprime[num]) {                    g.addEdge(boy[i], gir[j], 1);                    adore[boy[i]] = adore[gir[j]] = true;                }            }        }        int ans = wtf(n, B, k);        printf("%d\n", ans);    }    return 0;}

真的是佛了

匈牙利过了，拿前向星版的dinic死活re和wa，找了一年前的vector的dinic板子直接过了，一行一行对，没办法，拿vector的板子现场该前向星，还是狂wa

没错，下面这段代码wa到死，睿智了

#include <bits/stdc++.h>using namespace std;typedef long long LL;const int N = 3e3 + 7;const int INF = 0x3f3f3f3f;//-------------------------prime--------------------------------const int PRIME_N = 2e6 + 7;int prime[PRIME_N], primesize;bool isprime[PRIME_N + 7];void getprime(int listsize){    memset(isprime, 1, sizeof(isprime));    isprime[1] = false;    primesize = 0;    for (int i = 2; i < listsize; i++){        if (isprime[i]) prime[++primesize]=i;        for (int j = 1; j <= primesize && i*prime[j] < listsize; j++){            isprime[i*prime[j]] = false;            if (i % prime[j] == 0) break;        }    }}bool marry[N], adore[N];int arr[N], boy[N], gir[N], B, G;//------------------------Dinic-maxFlow--------------------------struct Dinic{    struct Edge{        int from, to, cap, flow, nxt;        Edge(){}        Edge(int u,int v,int c,int f, int n):from(u),to(v),cap(c),flow(f),nxt(n){}    } edges[N*N>>1];    int n, s, t, E, head[N];    bool vis[N]; //use when bfs    int d[N], cur[N]; //dist, now edge, use in dfs    inline void init(int _n, int _s, int _t){        n = _n; s = _s, t = _t; E = 0;        for (int i = 0; i <= n; i++) head[i] = -1;    }    inline void addEdge(int f, int t, int c){        edges[E] = Edge(f, t, c, 0, head[f]);        head[f] = E++;        edges[E] = Edge(t, f, 0, 0, head[t]);        head[t] = E++;    }    inline bool BFS(){        memset(vis, 0, sizeof(vis));        queue <int> Q;        d[s] = 0; vis[s] = 1;        for (Q.push(s); !Q.empty();){            int x = Q.front(); Q.pop();            for (int i = head[x]; ~i; i = edges[i].nxt){                Edge &e = edges[i];                if (vis[e.to] || e.cap <= e.flow) continue;                vis[e.to] = 1;                d[e.to] = d[x] + 1;                Q.push(e.to);            }        }        return vis[t];    }    inline int DFS(const int& x, int a){        //printf("dfs:%d,%d\n",x,a);        if (x == t || a == 0) return a;        int flow = 0, f;        for (int& i = cur[x]; ~i; i = edges[i].nxt){            Edge& e = edges[i];            if (d[x] + 1 != d[e.to]) continue;            if ((f = DFS(e.to,min(a,e.cap-e.flow))) <= 0)continue;            e.flow += f;            edges[i^1].flow-=f;//反向边            flow+=f; a-=f;            marry[x] = marry[e.to] = true;            if (a==0) break;        }        return flow;    }    inline int maxFlow(){return maxFlow(s,t);}    inline int maxFlow(const int& s, const int& t){        int flow = 0;        while (BFS()){            for (int i = 0; i <= n; i++) cur[i] = head[i];            int f = DFS(s, INF);            flow += f ;        }        return flow;    }} g;inline int wtf(const int &n, const int &B, const int &k){    memset(marry, false, sizeof marry);    int mat = g.maxFlow();    //printf("mat = %d\n", mat);    if (mat >= k) return k * 2;    int cnt1 = 0, idx = n; // number of one    for (int i = n; i >= 1; i--){        if (arr[i] != 1) {idx = i; break;}        if ((!marry[i])) cnt1++;        marry[i] = true;    }    //printf("cnt1 = %d\n", cnt1);    mat += cnt1 >> 1; // 剩下的1    if (mat >= k) return k*2;    int left = cnt1 & 1; // 单身狗们    for (int i = 1; i <= idx; i++){        left += (!marry[i]) & adore[i];    }    if (mat + left <= k) return mat*2+left;    return mat*2 + (k - mat);}int main(){    //freopen("in.txt", "r", stdin);    getprime(2e6 + 7);    int _;    scanf("%d", &_);    for (int n, k; _--;) {        scanf("%d%d", &n, &k);        for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);        sort(arr + 1, arr + n + 1, greater<int>());        B = G = 0;        g.init(n + 1, 0, n + 1);        for (int i = 1; i <= n; i++){            if (arr[i] & 1) {                boy[++B] = i;                g.addEdge(0, i, 1);            } else {                gir[++G] = i;                g.addEdge(i, n+1, 1);            }        }        memset(adore, false, sizeof adore);        for (int i = 1; i <= B; i++){            for (int j = 1; j <= G; j++){                int num = arr[boy[i]] + arr[gir[j]];                if (isprime[num]) {                    g.addEdge(boy[i], gir[j], 1);                    adore[boy[i]] = adore[gir[j]] = true;                }            }        }        int ans = wtf(n, B, k);        printf("%d\n", ans);    }    return 0;}