平时二十测

Posted 2021-01-16 edsheeran

 

 

 

 

题解；

第一题：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int M = 2e5 + 5;
const ll P=1e9 + 7;
ll ksm(ll a, ll b){
    ll ret=1;
    for(;b;b>>=1,a=a*a%P)
    if(b&1)ret=ret*a%P;
    return ret;
}
ll fac[M], vfac[M], A[M], B[M], l[M], t[M];
inline ll C(int n, int m){
    return fac[n] * vfac[m] % P * vfac[n-m]%P;
}

int main(){
    freopen("matrix.in","r",stdin);
    freopen("matrix.out","w",stdout);
    int n, a, b;
    scanf("%d%d%d",&n,&a,&b);
    fac[0]=1;vfac[0]=1;
    for(ll i=1;i<=(2*n);i++){
        fac[i]=fac[i-1]*i%P;
    }
    vfac[n]=ksm(fac[n], P-2);
    for(ll i=n-1;i;i--)vfac[i]=vfac[i+1]*(i+1)%P;
    
    for(int i=1;i<=n;i++)scanf("%lld",&l[i]);
    for(int i=1;i<=n;i++)scanf("%lld",&t[i]);
    if(n==1){printf("%lld
",t[1]);return 0;}
    ll t1=n-2,t2=0;
    ll ans=0;
    ll pa=1, pb=1, ap=ksm(a,n-1),bp=ksm(b,n-1);
    for (int i=n; i>1; i--,t1++,t2++,pa=1ll*pa*a%P,pb=1ll*pb*b%P)
        ans=((1ll*ans + 1ll*l[i]*ap%P*pb%P*C(t1,t2)%P)%P + 1ll*t[i]*bp%P*pa%P*C(t1,t2)%P)%P;
    printf("%lld
",ans);
}

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第二题：dp[i][j]表示可否拆成i个p,j个q，枚举每个数拆成多少个p,q贪心取，因为dp[i][j]可行, dp[i][k](k<j)也可行；复杂度（50*2000*2000*50）

#include<bits/stdc++.h>
using namespace std;
int a[55];
struct node{int a,b;}w[55][2005];
bool dp[2][2005][2005];
#define RG register
int main(){
    freopen("pq.in","r",stdin);
    freopen("pq.out","w",stdout);
    int n, p, q;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++)
        scanf("%d", &a[i]);
    scanf("%d%d",&p,&q);
    if(p < q)swap(p, q);
    int up1=0,up2=0;
    for(int i=1;i<=n;i++){
        int x=0;
        while(a[i]>=x*p){w[i][++w[i][0].a].a=x,w[i][w[i][0].a].b=(a[i]-x*p)/q;x++;}
        up1+=(a[i]/p);
        up2+=(a[i]/q);
    }
    memset(dp, 0, sizeof(dp));
    dp[0][0][0]=1;
    int ans=0,u=0;
    for(int i=1;i<=n;i++){
        u^=1;
        memset(dp[u], 0, sizeof(dp[u]));
        for(RG int h=up1;h>=0;h--)
        for(RG int z=up2;z>=0;z--)
        if(dp[u^1][h][z]){
            for(RG int j=1;j<=w[i][0].a;j++){
                {
                    dp[u][h+w[i][j].a][z+w[i][j].b] |= dp[u^1][h][z];
                //    printf("%d %d %d %d
",i,h,z,dp[u][h][z]);
                }
            }
        }
                
    }
    int c=p+q;
    for(int i = up1; i >= 0; i--)
        for(int j = i; j <= up2; j++)
            if(dp[u][i][j]) ans = max(ans, i*c);
    
    printf("%d
",ans);
}

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第三题：原题：https://www.cnblogs.com/EdSheeran/p/9740394.html

#include<bits/stdc++.h>
using namespace std;

const int M = 5e4, ME = 2e5 + 5;
#define ll long long
const ll inf = 1e9;
int tot = 1, h[M];ll dis[M];
bool inq[M];
queue<int>q;
struct edge{int v, nxt, w;}G[ME];
void add(int u, int v, int w){G[++tot].v=v,G[tot].nxt=h[u],G[tot].w=w,h[u]=tot;}

void SPFA(){
    memset(inq, 0, sizeof(inq));
    while(!q.empty()){
        int u=q.front();q.pop();inq[u]=0;
        for(int i=h[u];i;i=G[i].nxt){
            int v=G[i].v;
            if(v!=1&&dis[v]>dis[u]+G[i].w){
                dis[v]=dis[u]+G[i].w;
                //printf("%d %d %lld
",u,v,dis[v]);
                if(!inq[v])inq[v]=1,q.push(v);
            }
        }
    }
}
int read(){
    int x=0,f=1;char c=getchar();
    while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();}
    return x*=f;
}


int main(){
    freopen("graph.in","r",stdin);
    freopen("graph.out","w",stdout);
    int n = read(), m = read();
    for(int i = 1; i <= m; i++){
        int u = read(), v = read(), w1 = read(), w2 = read();
        add(u, v, w1); add(v, u, w2);
    }
    ll ans = inf;
    for(int bit = 1; bit <= 18; bit++){
        memset(dis, 127, sizeof(dis));
        for(int i = h[1]; i; i = G[i].nxt){
            int v = G[i].v;
            if(v & (1<<(bit-1))) 
            q.push(v), dis[v] = min(dis[v], 1LL*G[i].w), inq[v] = 1;
            
        }
        SPFA();
        for(int i = h[1]; i; i = G[i].nxt){
            int v = G[i].v;
            if((v & (1<<(bit-1))) == 0) {
                ans = min(ans, 1LL*G[i^1].w + dis[v]);
            }
        }
        memset(dis, 127, sizeof(dis));
        for(int i = h[1]; i; i = G[i].nxt){
            int v = G[i].v;
            if((v & (1<<(bit-1))) == 0) 
            q.push(v), dis[v] = min(dis[v], 1LL*G[i].w), inq[v] = 1;
            
        }
        SPFA();
        for(int i = h[1]; i; i = G[i].nxt){
            int v = G[i].v;
             if(v & (1<<(bit-1))){
                ans = min(ans, 1LL*G[i^1].w + dis[v]);
            }
        }
    }
    printf("%lld
", ans);
}

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