Codeforces Round #675 (Div. 2)

http://codeforces.com/contest/1422/problem/D

D. Returning Home

一个n*n的图

有些点就是可以瞬移  和象棋里车的规则一样，给你一些点，只要你在横纵坐标 就可以瞬移过去

考虑对x，y的列和行建图就可以了

 

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define inf64 0x3f3f3f3f3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn = 1e6+10;
const int mod = 1e9+7;
int head[maxn],nxt[maxn<<1],to[maxn<<1],w[maxn<<1],cnt;
void add(int u,int v,ll c){
    // printf("u = %d v = %d c = %lld\n", u,v,c);
    ++cnt,to[cnt]=v,w[cnt]=c,nxt[cnt]=head[u],head[u]=cnt;
    ++cnt,to[cnt]=u,w[cnt]=c,nxt[cnt]=head[v],head[v]=cnt;
}

struct heapnode{
    ll d,u;
    heapnode(ll d=0,ll u=0) : d(d),u(u) {}
    bool operator<(const heapnode &a) const{
        return a.d<d;
    }
};
ll d[maxn];
bool vis[maxn];
priority_queue<heapnode>que;
void dijkstra(int s,int n){
    while(!que.empty()) que.pop();
    for(int i=0;i<=n;i++) d[i]=inf64,vis[i] = false;
    d[s]=0,que.push(heapnode(0,s));
    while(!que.empty()){
        heapnode x=que.top();que.pop();
        int u=x.u;
        if(vis[u]) continue;
        vis[u]=true;
        for(int i=head[u];i;i=nxt[i]){
            int v = to[i];
            if(d[v]>d[u]+w[i]){
                d[v]=d[u]+w[i];
                que.push(heapnode(d[v],v));
            }
        }
    }
}

int a[maxn],b[maxn],x[maxn],y[maxn];
int main(){
    int n,m,sx,sy,gx,gy;
    scanf("%d%d",&n,&m);
    scanf("%d%d%d%d",&sx,&sy,&gx,&gy);
    for(int i=1;i<=m;i++){
        scanf("%d%d",&x[i],&y[i]);
        a[i] = x[i],b[i] = y[i];
    }
    sort(a+1,a+1+m);
    sort(b+1,b+1+m);
    add(0,3*m+1,abs(sx-gx)+abs(sy-gy));
    for(int i=2;i<=m;i++){
        add(i+m,i+m-1,abs(a[i]-a[i-1]));
        add(i+2*m,i+2*m-1,abs(b[i]-b[i-1]));
    }
    for(int i=1;i<=m;i++){
        add(0,i,min(abs(sx-x[i]),abs(sy-y[i])));
        add(i,3*m+1,abs(gx-x[i])+abs(gy-y[i]));
        int pos1 = lower_bound(a+1,a+1+m,x[i])-a;
        add(i,pos1+m,0);
        int pos2 = lower_bound(b+1,b+1+m,y[i])-b;
        add(i,pos2+2*m,0);
    }
    dijkstra(0,3*m+1);
    printf("%lld\n", d[3*m+1]);
    return 0;
}

C. Bargain

考虑 前缀对应后面多个后缀就可以了 求个后缀的后缀和

#include<bits/stdc++.h>
#define fi first
#define se second
#define io std::ios::sync_with_stdio(false)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n){ll r=1%P;for (a%=P; n; a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
const int mod=1e9+7;
const int maxn=1e5+10;
ll pre[maxn],pree[maxn],suf[maxn],suff[maxn];
char a[maxn];
int main()
{


    cin>>a+1;
    int n=strlen(a+1);
    ll ans=0;
    for(int i=1;i<=n;i++)
    {
        pre[i]=pre[i-1]*10+a[i]-'0';
        pre[i]%=mod;
        if(i!=n)
        ans+=pre[i];
        ans%=mod;
        pree[i]=pree[i-1]+pre[i];
        pree[i]%=mod;
    }
    ll cnt=1;
   // cout<<ans<<endl;
    for(int i=n;i>=1;i--)
    {
        suf[i]=suf[i+1]+cnt*(a[i]-'0');
        suf[i]%=mod;
        if(i!=1)
        ans+=suf[i];
        ans%=mod;
        suff[i]=suff[i+1]+suf[i];
        suff[i]%=mod;
        cnt*=10;
        cnt%=mod;
    }
    //cout<<ans<<endl;
    cnt=10;
    int num=1;
    for(int i=n-2;i>=1;i--)
      {
          ans+=pre[i]*cnt+suff[i+2];
          ans%=mod;
          num++;
          cnt+=qpow(10,num);
          cnt%=mod;
      }
      cout<<ans<<endl;

}