2014 8.8 第9场个人 排位

UVA 12657 Boxes in a Line

模拟

FZU 1876 JinYueTuan 不说话

view code#include <iostream>



using namespace std;



int main()

{

    long long n,m,p,ans=1;

    while(cin>>n>>m>>p){

        ans=1;

        for(int i=n;i<n+m;i++){

            ans=ans*i%p;

            if(ans==0)break;

        }

        cout<<ans<<endl;

    }

    return 0;

}

ZOJ 3684 Destroy

view code#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <queue>

using namespace std;

typedef long long ll;

const int INF = 0x3f3f3f3f;

const int N = 10010;

int _, cas=1, n, m,pre[N], ans=0;

int in[N],dp[N<<1];

bool vis[N<<1];



struct edge

{

    int u, v, w, p, next;

    edge() {}

    edge(int u, int v, int w, int p, int next):u(u),v(v),w(w),p(p),next(next) {}

}e[N<<1];

int ecnt;



int dfs(int u, int h, int p)

{

    int ans = 0;

    for(int i=pre[u]; ~i; i=e[i].next)

    {

        int v = e[i].v;

        if(v==p) continue;

        if(!vis[i]) dp[i] = dfs(v, e[i].w, u),vis[i] = 1;

        ans = max(dp[i], ans);

    }

    return h+ans;

}



void addedge(int u, int v, int w, int p)

{

    e[ecnt] = edge(u, v, w, p, pre[u]);

    pre[u] = ecnt++;

    e[ecnt] = edge(v, u, w, p, pre[v]);

    pre[v] = ecnt++;

}



//bool dfs2(int u, int lim, int p)

//{

// if(in[u]==1) return 0;

// for(int i=pre[u]; ~i; i=e[i].next)

// {

// int v = e[i].v;

// if(v==p || e[i].p<=lim) continue;

// if(!dfs2(v, lim, u)) return 0;

// }

// return 1;

//}



void dfs3(int u, int h, int p)

{

    if(in[u]==1)

    {

        ans = max(ans, h);

        return ;

    }

    for(int i=pre[u]; ~i; i=e[i].next)

    {

        int v = e[i].v;

        if(v==p) continue;

        dfs3(v, min(h,e[i].p), u);

    }

}



void solve()

{



    memset(pre, -1, sizeof(pre));

    memset(in, 0, sizeof(in));

    ecnt = 0;

    int u, v, w, p, l = 0, r = 0;

    for(int i=1; i<n; i++)

    {

        scanf("%d%d%d%d", &u, &v, &w, &p);

        addedge(u, v, w, p);

        in[u]++;

        in[v]++;

// r = max(p, r);

    }



    int s=0, Min=INF, tmp;

    memset(vis, 0, sizeof(vis));

    for(int i=1; i<=n; i++)

    {

        tmp = dfs(i, 0, 0);

        if(tmp<Min) s = i, Min=tmp;

    }

    ans = 0;

    dfs3(s, INF, 0);

// while(l<=r)

// {

// int mid = (l+r)>>1;

// if(dfs2(s, mid, 0)) ans =mid, r=mid-1;

// else l = mid+1;

// }

    printf("%d\n", ans);

}



int main()

{

// freopen("in.txt", "r", stdin);

    while(scanf("%d", &n)>0)  solve();

    return 0;

}

ZOJ 3676 Edward's Cola Plan （总是看英文看很久，囧）

view code#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

typedef long long ll;

const int INF = 0x3f3f3f3f;

const int N = 100010;

int _, cas=1, n, m;

int suma[N], sumb[N], ans[10010];



struct edge

{

    int a, b, c;

    edge() {}

    edge(int a, int b, int c):a(a),b(b),c(c) {}

    bool operator < (const edge &o) const{

        return c>o.c;

    }

}e[N];



struct node

{

    int v, id;

    bool operator < (const node &o) const{

        return v>o.v;

    }

}p[10010];



void solve()

{

    int a, b;

    for(int i=1; i<=n; i++){

        scanf("%d%d",&a, &b);

        e[i] = edge(a,b, b-a);

    }

    sort(e+1, e+1+n);

    for(int i=1; i<=n; i++)

    {

        suma[i] = suma[i-1] + e[i].a;

        sumb[i] = sumb[i-1] + e[i].b;

    }

    for(int i=0; i<m; i++)

    {

        scanf("%d", &p[i].v);

        p[i].id = i;

    }

    sort(p, p+m);



    int j = 1, M;

    for(int i=0; i<m; i++)

    {

        M = p[i].v;

        while(j<=n && e[j].c>M) j++;

        ans[p[i].id] = sumb[j-1]-M*(j-1)+suma[n]-suma[j-1];

    }

    for(int i=0; i<m; i++)

    {

        printf("%d\n", ans[i]);

    }

}



int main()

{

// freopen("in", "r", stdin);

    while(scanf("%d%d", &n,&m)>0) solve();

    return 0;

}

ZOJ 2899 Hangzhou Tour

 

ZOJ 3724 Icecream 线段树，离线，挺经典的

sum[n]表示距离的前缀和

情况1：

如果是询问从u到v的最短距离(u<v)，(a到b是一条捷径(a<b)，长度为len)

如果不走捷径dis1 = sumv – sumu;

如果走捷径 dis2 = sumv – sumu +len - (sumb-suma)。

走不走捷径就看len-(sumb-suma)是否小于0,(等于0,走不走无所谓)

对于多个捷径，且走捷径的情况，走len-(sumb-suma)最小的那个捷径

情况2：

如果是询问从u到v的最短距离（u>v），(a到b是一条捷径(a>b)，长度为len)

这种情况必走捷径：

走捷径dis = suma – sumu + sumv – sumb + len;

对于一个询问sumv-sumu是固定的，关键要求走的捷径使suma-sumb+len最小

 

至于那些最小值就是用线段树去维护。

---

view code#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <queue>

using namespace std;

typedef long long ll;

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

const ll INF = 1LL<<60;

const int N = 100010;

int n, m, T;

ll Min[N<<2], sum[N], ans[N<<1];



struct seg

{

    int l, r; ll w;

    seg() {}

    seg(int l, int r, ll w):l(l),r(r),w(w) {}

    bool operator < (const seg &o) const{

        return r<o.r;

    }

}e1[N<<1], e2[N<<2], ask1[N<<2], ask2[N<<2];

int ecnt1, ecnt2, acnt1, acnt2;



bool cmp(const seg a, const seg b)

{

    return a.l<b.l;

}



void Up(int rt)

{

    Min[rt] = min(Min[rt<<1], Min[rt<<1|1]);

}



void build(int l, int r, int rt)

{

    Min[rt] = INF;

    if(l==r) return ;

    int m = (l+r)>>1;

    build(lson);

    build(rson);

}



void update(int p, ll c, int l, int r, int rt)

{

    if(l==r)

    {

        Min[rt] = min(Min[rt], c);

        return ;

    }

    int m = (l+r)>>1;

    if(p<=m) update(p, c, lson);

    else update(p, c, rson);

    Up(rt);

}



ll query(int L, int R, int l, int r, int rt)

{

    if(L<=l && R>=r) return Min[rt];

    int m = (l+r)>>1;

    ll ans = INF;

    if(L<=m) ans = min(ans, query(L, R, lson));

    if(R>m)  ans = min(ans, query(L, R, rson));

    return ans;

}



void solve()

{

    int a, b, c;

    for(int i=2; i<=n; i++)

    {

        scanf("%d", &a);

        sum[i] = sum[i-1] + a;

    }



    ecnt1 = ecnt2 = acnt1 = acnt2 = 0;

    while(m--)

    {

        scanf("%d%d%d", &a, &b, &c);

        if(a<b)

        {

            if(sum[a] - sum[b] + c>0) continue;

            e1[ecnt1++] = seg(a, b, c-(sum[b]-sum[a]));

        }

        else e2[ecnt2++] = seg(b, a, (sum[a]-sum[b]+c));

    }



    scanf("%d", &T);

    for(int i=0; i<T; i++)

    {

        scanf("%d%d", &a, &b);

        if(a<b) ask1[acnt1++] = seg(a, b, i);

        else if(a>b)ask2[acnt2++] = seg(b, a, i);

        else ans[i] = 0;

    }



    sort(ask1, ask1+acnt1);

    sort(e1, e1+ecnt1);

    int i=0, j=0;

    build(1, n, 1);

    for(; i<acnt1; i++)

    {

        while(j<ecnt1 && e1[j].r<=ask1[i].r)

        {

            update(e1[j].l, e1[j].w, 1, n, 1);

            j++;

        }

        ll t = min((ll)0,query(ask1[i].l, ask1[i].r, 1, n, 1));

        ans[ask1[i].w] = sum[ask1[i].r] - sum[ask1[i].l] + t;

    }

    sort(ask2, ask2+acnt2, cmp);

    sort(e2, e2+ecnt2, cmp);



    build(1, n, 1);

    for(i = 0, j = 0; i<acnt2; i++)

    {

        while(j<ecnt2 && e2[j].l<=ask2[i].l)

        {

            update(e2[j].r, e2[j].w, 1, n, 1);

            j++;

        }

        ans[ask2[i].w] =sum[ask2[i].l]-sum[ask2[i].r]+query( ask2[i].r, n, 1, n, 1);

// printf("(%d, %d)first : %I64d\n",ask2[i].l,ask2[i].r,ans[ask2[i].w]);

    }

    for(int i=0; i<T; i++)

    {

        printf("%lld\n", ans[i]);

    }

}



int main()

{

// freopen("in.txt", "r", stdin);

    while(scanf("%d%d", &n, &m)>0) solve();

    return 0;

}