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）
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#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）
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#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\n", 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
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#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\n", ans);
}
}
并查集
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#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\n", 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
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#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("\n");
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）
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#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\n", flow);
if(flow!=n) puts("-1");
else printf("%d\n", cost);
}
return 0;
}