20180908沈阳网络赛题解

目录

• 20180910沈阳网络赛题解
  • D. Made In Heaven
    • MEANING
    • SOLUTION
    • CODE
  • F. Fantastic Graph
    • MEANING
    • SOLUTION
    • CODE
  • G. Spare Tire
    • CODE
  • I. Lattice's basics in digital electronics
    • CODE
  • K. Supreme Number

20180910沈阳网络赛题解

D. Made In Heaven

MEANING

给出一张有向图，询问S到E的第K短路的长度是否大于T。

SOLUTION

和poj2449一样，A*算法。

CODE

#define IN_PC() freopen("C:\Users\hz\Desktop\in.txt","r",stdin)
#define IN_LB() freopen("C:\Users\acm2018\Desktop\in.txt","r",stdin)
#define OUT_PC() freopen("C:\Users\hz\Desktop\out.txt","w",stdout)
#define OUT_LB() freopen("C:\Users\acm2018\Desktop\out.txt","w",stdout)
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 1e3 + 5;
const int MAXE = 1e4 + 5;
const int INF = 0x3f3f3f3f;

struct Graph {
    struct edge {
        int v,w,nex;
    } ed[MAXE];
    int tot,head[MAXN];
    void addedge(int u,int v,int w) {
        tot++;
        ed[tot].v = v;
        ed[tot].w = w;
        ed[tot].nex = head[u];
        head[u] = tot;
    }
    void init() {
        tot = 0;
        memset(head,0,sizeof head);
    }
} G1,G2;

struct node {
    int ind,dis;
};
bool operator <(node a,node b) {
    return a.dis>b.dis;
}
int d[MAXN];
bool vis[MAXN];
priority_queue<node> q;
void dijkstra(int st,Graph G) {
    memset(d,0x3f,sizeof d);
    memset(vis,0,sizeof vis);
    d[st] = 0;
    q.push({st,0});
    while(q.size()) {
        int u = q.top().ind;
        q.pop();
        if(vis[u])
            continue;
        vis[u] = 1;
        for(int i = G.head[u]; i; i=G.ed[i].nex) {
            int v = G.ed[i].v,w = G.ed[i].w;
            if(d[v]>d[u]+w) {
                d[v] = d[u]+w;
                q.push({v,d[v]});
            }
        }
    }
}

struct anode {
    int to;
    int f,g;
    bool operator < (const anode &t) const {
        if(t.f==f)
            return t.g < g;
        return t.f < f;
    }
};
int Astar(int st,int et,int k) {
    int cnt = 0;
    priority_queue<anode> pq;
    if(st==et)
        k++;
    if(d[st] == INF)
        return -1;
    pq.push({st,d[st],0});
    while(pq.size()){
        anode u = pq.top();
        if(u.g>100000000)return -1;
        pq.pop();
        if(u.to==et)cnt++;
        if(cnt==k)return u.g;
        for(int i = G1.head[u.to];i;i=G1.ed[i].nex){
            anode v;
            v.to = G1.ed[i].v;
            v.g = u.g + G1.ed[i].w;
            v.f = v.g + d[v.to];
            pq.push(v);
        }
    }
    return -1;
}

int main() {
//    IN_LB();
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF) {
        int st,et,k,t;
        scanf("%d%d%d%d",&st,&et,&k,&t);
        G1.init();
        G2.init();
        for(int i=0; i<m; i++) {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            G1.addedge(u,v,w);
            G2.addedge(v,u,w);
        }
        dijkstra(et,G2);//在反图上跑Dijkstra
        int ans = Astar(st,et,k);//在正图上跑Astar
        if(ans!=-1&&ans<=t)printf("yareyaredawa
");
        else printf("Whitesnake!
");
    }
    return 0;
}

F. Fantastic Graph

MEANING

给一个二分图，询问能否找到一种匹配，使得每个点相连的匹配边数在[L,R]内。

SOLUTION

解法一（官方题解）：有源汇的上下界网络流的可行流问题。
解法二：Dinic算法跑二分图多重匹配。先令左部图的点取流量上界（超级源点与左部图各点连一条容量为R的边），右部图的点取流量下界（超级汇点与右部图各点连一条容量为L的边）,跑最大流。
然后令右部图的点取流量上界（超级源点与右部图各点连一条容量为R的边），左部图的点取流量下界（超级汇点与左部图各点连一条容量为L的边），再跑一边最大流。如果两次与汇点相连的边都跑满流量，说明可行，否则不可行。

CODE

G. Spare Tire

CODE

队友代码

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
const int maxn = 10050;

ll prime[maxn];
ll isCon[maxn];
ll cnt = 0;

void eurla(){
    for (int i=2;i<maxn;i++){
        if (isCon[i]==0) {
            prime[cnt++] = i;
          //  cout<<i<<endl;
        }
        for (int j=0;j<cnt&&i*prime[j]<maxn;j++){
             isCon[ i*prime[j] ] = 1;
             if (i%prime[j]==0) break;
        }
    }
}


ll qpow(ll a,ll x){
    ll ret = 1;
    while(x){
        if (x&1) ret=ret*a%mod;
        a=a*a%mod;
        x>>=1;
    }
    return ret;
}
ll a[15];

int main()
{
    eurla();
    ll inv,inv2;
    inv = qpow(6,mod-2);
    inv2 = qpow(2,mod-2);
    ll k,m;
    while(~scanf("%lld %lld",&k,&m))
    {
        ll p = m;
        ll cntt=0;
        for(int i=0;i<cnt;i++)
        {
            if(p<prime[i])
                break;
            if (p%prime[i]==0){
                    a[cntt++] = prime[i];
               while (p%prime[i]==0)
                p = p/prime[i];
            }

        }
        if (p!=1) a[cntt++] = p;
//        for (int i=0;i<cntt;i++){
//            cout<<a[i]<<" ";
//        }cout<<endl;
        ll ans=0;
        for(int i=1;i<(1<<cntt);i++)
        {
            ll x=0,y=1;
            for(int j=0;j<cntt;j++)
            {
                if((i>>j)&1)
                {
                    x++;
                    y = y*a[j]%mod;
                }
            }
           // printf("%lld
",y);
           ll n = k/y;
           if(x%2)
                ans = (ans+y*y%mod*n%mod*(n+1)%mod*(2*n+1)%mod*inv%mod+y*n*(n+1)%mod*inv2%mod+mod)%mod;
            else
                ans = (ans-y*y%mod*n%mod*(n+1)%mod*(2*n+1)%mod*inv%mod-y*n*(n+1)%mod*inv2%mod+mod)%mod;
        }
        ll s;
        s = k*(k+1)%mod*(2*k+1)%mod*inv%mod+k*(k+1)%mod*inv2%mod;
      //  printf("%lld
",ans);
        ans = (s-ans+mod)%mod;
        printf("%lld
",ans);

    }
    return 0;
}

I. Lattice's basics in digital electronics

按题意模拟即可。

CODE

队友代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pll;
const ll maxn = 2000006;
const ll mod = 1000000007;
const ll trieNum = 5010;

struct Decoder {
    struct Trie {
        int isEnd;
        int nxt[2];
        int val;
    } trie[trieNum];
    int root = 0;
    int cnt = 0;
    void init() {
        root =cnt = 0;
        memset(trie,0,sizeof trie );
    }
    void insert(char s[],int val) {
        int p = root;
        int i=0;
        for (; s[i]; i++) {
            int j = s[i]-'0';
            if (trie[p].nxt[j]==0 ) {
                trie[p].nxt[j] = ++cnt;
            }
            p = trie[p].nxt[j];
        }
        trie[p].isEnd = 1;
        trie[p].val = val ;
    }
    void decode(char s[],char d[]) {
        int i=0;
        int tot = 0;
        int p = root;
        for (; s[i]; i++) {
            p = trie[p].nxt[s[i]-'0'];
            if (trie[p].isEnd==1) {
                d[tot++] = trie[p].val ;
                p=root;
            }
        }
        d[tot] = '';
    }
} decoder;

void x2b(char s[],char d[],int len) {
    for (int i=0; i<len; i++) {
        int v;
        if (isdigit(s[i]))
            v = s[i]-'0';
        else
            v = s[i]-'A'+10;
        for (int j=0; j<4; j++) {
            d[i*4 + 3 - j ] = ((v>>j)&1)+'0' ;
        }
    }
    d[len*4] = '';
}

void detect(char s[],char d[],int len) {
    int tot = 0;
    for (int i=0; (i+1)*9<=len; i++) {
        int k = 0;
        for (int j=0; j<9; j++) {
            if (s[ i*9 + j ]== '1' )
                k++;
        }
        if (k%2) {
            for (int j=0; j<8; j++) {
                d[tot++] = s[i*9+j];
            }
        }
    }
    d[tot]='';
}

char s1[maxn],s2[maxn],s3[maxn],tmp[200];

int main() {
    // freopen("in.txt","r",stdin);
    int t;
    scanf("%d",&t);
    while(t--) {
        int n,m;
        scanf("%d%d",&m,&n);
        decoder.init();
        for (int i=1; i<=n; i++) {
            int val;
            scanf("%d",&val);
            scanf("%s",tmp);
            decoder.insert(tmp,val);
        }
        scanf("%s",s1);
        int len = strlen(s1);
        x2b(s1,s2,len);
        detect(s2,s1,len*4);
        decoder.decode(s1,s3);
        for (int i=0; i<m; i++) {
            printf("%c",s3[i]);
        }
        printf("
");
    }
    return 0;
}

K. Supreme Number

签到题。就看谁找素数速度更快了。现在纸上枚举1，2，3，5，7的可能组合，可以肯定不会超过四位。对于无法判断是否素数的，写程序暴力判即可。