2014 8.6 第7场个人 排位

这次比赛有显示了自己的不足，之所以是又，是因为基本上每次都暴露了自己很水，囧。

剩下的时间努力刷题，加油

CodeForces 287B Pipeline （二分加贪心）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
ll n, m;

ll solve()
{
    if(n==1) return 0;
    ll sum = (1+m)*m/2-(m-2)-1;
    if(sum<n) return -1;
    ll l = 2, r = m, ans=0;
    while(l<=r)
    {
        int mid =(l+r)>>1;
        if((m-mid+1)*(mid+m)/2-(m-mid)>=n) ans = m-mid+1, l = mid+1;
        else r = mid-1;
    }
    return ans;
}

int main()
{
    while(cin>>n>>m)
        cout<<solve()<<endl;
    return 0;
}

HDU 4366 Successor （单点更新最值，区间查询最值，经典的地方在于将树转化为一个线性序列）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long ll;
const int N = 5e4+10;
int _, n, m, pre[N], L[N], R[N], ans[N], Max[N<<2];
map<int , int> ma;

struct edge
{
    int v, next;
    edge() {}
    edge(int v, int next):v(v),next(next) {}
}e[N];
int ecnt, tot;

struct people
{
    int id, loy, abi;
    bool operator < (const people &o) const{
        return abi>o.abi;
    }
}man[N];

void addedge(int u, int v)
{
    e[ecnt] = edge(v, pre[u]);
    pre[u] = ecnt++;
}

void dfs(int u)//将树转化为一个线性序列
{
    L[u] = tot++;
    for(int i=pre[u]; ~i; i=e[i].next)
        dfs(e[i].v);
    R[u] = tot;
}

void update(int p, int c, int l, int r, int rt)
{
    if(l==r) {Max[rt] = c; return; }
    int m = (l+r)>>1;
    if(p<=m) update(p, c, lson);
    else update(p, c, rson);
    Max[rt] = max(Max[rt<<1], Max[rt<<1|1]);
}

int query(int L, int R, int l, int r, int rt)
{
    if(L<=l && R>=r) return Max[rt];
    int ans = -1, m = (l+r)>>1;
    if(L<=m) ans = max(ans, query(L,R,lson));
    if(R>m) ans = max(ans, query(L,R,rson));
    return ans;
}

void init()
{
    memset(L, 0, sizeof(L));
    memset(R, 0, sizeof(R));
    memset(pre, -1, sizeof(pre));
    memset(Max, -1, sizeof(Max));
    memset(ans, -1, sizeof(ans));
    tot = ecnt = 0; ma.clear();
    ma[-1] = -1;
}

void solve()
{
    init();
    scanf("%d%d", &n, &m);

    int fa, loy;
    for(int i=1; i<n; i++)
    {
        scanf("%d%d%d", &fa, &man[i].loy, &man[i].abi);
        man[i].id = i;
        addedge(fa, i);
        ma[man[i].loy] = i;
    }
    dfs(0);
    sort(man+1, man+n);

    int j, id;
    for(int i=1; i<n; i = j)
    {
        j = i;
        while(j<n && man[j].abi==man[i].abi)
        {
            id = man[j].id;
            if(L[id]+1<=R[id]-1)
               loy = query(L[id]+1, R[id]-1, 0, tot-1, 1);
            else loy = -1;
            ans[id] = ma[loy];
            j++;
        }
        j = i;
        while(j<n && man[j].abi==man[i].abi)
        {
            id = man[j].id;
            update(L[id], man[j].loy, 0, tot-1, 1);
            j++;
        }
    }

    while(m--)
    {
        scanf("%d", &id);
        printf("%d
", ans[id]);
    }
}

int main()
{
//    freopen("in.txt", "r", stdin);
    scanf("%d", &_);
    while(_--) solve();
    return 0;
}

HDU 3047  Zjnu Stadium（带权值的并查集）**（比赛是这道题竟然做不出，囧，只怪自己太弱）

先来一个类似模拟的做法：比赛的时候没过，汗

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <map>
#include <string>
#define REP(i,n) for(int i=0; i<(n); i++)
#define FOR(i,s,t) for(int i=(s); i<=(t); i++)
using namespace std;
typedef long long ll;
const int INF = 1<<30;
const int N = 50010;
//const int mod = 300;
int id[N], vis[N], n, m;
vector<int > vec[N];

void add(int cnt, int w, int now, int disu, int disv)
{
    int siz = vec[cnt].size(), v;
    for(int i=0; i<siz; i++)
    {
        v = vec[cnt][i];
        id[v] += disu+w-disv;
        vis[v] = now;
        vec[now].push_back(v);
    }
    vec[cnt].clear();
}

int main()
{
//    freopen("in.txt", "r", stdin);
    while(scanf("%d%d", &n, &m)>0)
    {
        int ans = 0, u, v, w;
        memset(vis, 0, sizeof(vis));
        for(int i=0; i<=n; i++) vec[i].clear();
        int cnt = 0;
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d%d", &u, &v, &w);
            if(!vis[u] && !vis[v])
            {
                id[u] = 0, id[v] = w;
                vis[u] = vis[v] = ++cnt;
                vec[cnt].push_back(u);
                vec[cnt].push_back(v);
            }
            else if(vis[u] == vis[v])
            {
                if(id[v]-id[u]!=w) ans++;
            }
            else if(vis[u] && vis[v])
            {
                add(vis[v], w, vis[u], id[u], id[v]);
            }
            else if(vis[u])
            {
                id[v] = id[u]+w;
                vis[v] = vis[u];
                vec[vis[u]].push_back(v);
            }
            else if(vis[v])
            {
                id[u] = id[v]-w;
                vis[u] = vis[v];
                vec[vis[v]].push_back(u);
            }
        }
        printf("%d
", ans);
    }
}

并查集

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
typedef long long ll;
const int N = 5e4+10;
int fa[N], dis[N], n, m;

int find(int x)
{
    if(x==fa[x]) return fa[x];
    int t = fa[x];
    fa[x] = find(fa[x]);
    dis[x] += dis[t];
    return fa[x];
}

int main()
{
//    freopen("in.txt", "r", stdin);
    while(scanf("%d%d", &n, &m)>0)
    {
        for(int i=1; i<=n; i++) fa[i] = i, dis[i] = 0;
        int u, v, w, fu, fv, ans = 0;
        for(int i=0; i<m; i++)
        {
            scanf("%d%d%d", &u, &v, &w);
            fu = find(u), fv = find(v);
            if(fu==fv)
            {
                if(dis[u]+w!=dis[v]) ans++;
            }
            else
            {
                fa[fv] = fu;
                dis[fv] = w-dis[v]+dis[u];
            }
        }
        printf("%d
", ans);
    }
    return 0;
}

HDU 4886 TIANKENG’s restaurant(Ⅱ)（据说很暴力的做法都能过trie，平时很少接触字符串类的问题）(not yet）

8进制，队友代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int maxn=1000005;
int vis[maxn<<2];
char str[maxn];
int frc[10];
int n;

bool solve(int k){
    memset(vis,0,sizeof vis);
    int tmp=0;
    for(int i=0;i<k;i++)tmp=tmp*8+str[i]-'A';
    vis[tmp]=1;
    for(int i=k;i<n;i++){
        tmp=tmp%frc[k-1];
        tmp=tmp*8+str[i]-'A';
        vis[tmp]=1;
    }
    int x=0;
    while(vis[x])x++;
    if(x<frc[k]){
        for(int i=k;i>=1;i--){
            printf("%c",x/frc[i-1]+'A');
            x%=frc[i-1];
        }
        printf("
");
        return 1;
    }
    return 0;
}

int main()
{
//    freopen("in","r",stdin);
    frc[0]=1;
    for(int i=1;i<=8;i++)frc[i]=frc[i-1]*8;
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%s",str);
        n=strlen(str);
        for(int i=1;!solve(i);i++);
    }
    return 0;
}

UVA 12672  Eleven(如果一个数的奇数位之和减去偶数位之和能被11整除，该数就能被11整除)（not yet）

HDU 1853 Cyclic Tour（拆点，最大流）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
const int INF = 1<<30;
const int N = 210;
int n, m, p[N], d[N], pre[N];
int s, t;
bool inq[N];

struct edge
{
    int u, v, cap, cost, next;
    edge() {}
    edge(int u, int v, int cap, int cost, int next):u(u), v(v),cap(cap),cost(cost),next(next) {}
}e[N*N*4];
int ecnt;

void addedge(int u, int v, int w, int cost)
{
    e[ecnt] = edge(u, v, w, cost, pre[u]);
    pre[u] = ecnt++;
    e[ecnt] = edge(v, u, 0, -cost, pre[v]);
    pre[v] = ecnt++;
}

bool BellmanFord(int &flow, int &cost)
{
    for(int i=s; i<=t; i++) d[i] = INF;
    memset(inq, 0, sizeof(inq));
    d[s] = 0; p[s] = 0;
    int minflow = INF;

    queue<int >q;
    q.push(s);
    while(!q.empty())
    {
        int u = q.front(); q.pop();
        inq[u] = 0;
        for(int i = pre[u]; ~i; i = e[i].next)
        {
            int v = e[i].v;
            if(e[i].cap>0 && d[v]>d[u]+e[i].cost)
            {
                d[v] = d[u]+e[i].cost;
                p[v] = i;
                minflow = min(minflow, e[i].cap);
                if(!inq[v]) q.push(v), inq[v] = 1;
            }
        }
    }

    if(d[t] == INF) return false;
    flow += minflow;
    cost += d[t]*minflow;
    int u = t;
    while(u != s)
    {
        e[p[u]].cap -= minflow;
        e[p[u]^1].cap += minflow;
        u = e[p[u]].u;
    }
    return true;
}

int Mincost(int &flow)
{
    int cost=0;
    while(BellmanFord(flow, cost));
    return cost;
}

int main()
{
//    freopen("in.txt", "r", stdin);
    while(scanf("%d%d", &n, &m)>0)
    {
        memset(pre, -1, sizeof(pre));
        ecnt = 0, s = 0, t = n<<1|1;
        int u, v, w;
        for(int i=1; i<=n; i++) addedge(s,i,1,0), addedge(i+n,t,1,0);
        for(int i=0; i<m; i++)
        {
            scanf("%d%d%d", &u, &v, &w);
            addedge(u, v+n, 1, w);
        }

        int flow=0;
        int cost=Mincost(flow);
//        printf("mainflow = %d
", flow);
        if(flow!=n) puts("-1");
        else printf("%d
", cost);
    }
    return 0;
}