2015 Multi-University Training Contest 1 题解 BY ME
多校第一场结束了，先来没事整理了一下代码
http://blog.sina.com.cn/s/blog_15139f1a10102vnx5.html （这是题解，没有代码的题解）
下面是我根据题解写出的代码
第一题 hdu 5288 OO’s Sequence
#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <vector>
#include <sstream>
#include <cstdlib>
#include <complex>
#include <cstring>
#include <iostream>
#include <algorithm>
#define REP(i,N) for (int i = 0;i < (N);i++)
#define REP_1(i,N) for (int i = 1;i < (N);i++)
#define REP_2(i,be,en) for (int i = (be);i < (en);i++)
#define DWN(i,N) for (int i = (N);i >= 0;i--)
#define DWN_1(i,N) for (int i = (N);i >= 1;i--)
#define DWN_2(i,en,be) for (int i = (en);i >= (be);i--)
#define FR(N) freopen((N),"r",stdin)
#define FW(N) freopen((N),"w",stdout)
#define MAXN 2010
#define GETS(ch) fgets((ch),MAXN,stdin)
#define INF 0x3f3f3f3f
#define MOD 1000000007
using namespace std;
typedef long long LL;
typedef LL ll;
typedef map<int,LL> MINT;
typedef map<char,int> MCH;
typedef map<string,int> MSTR;
typedef vector<int> VINT;
typedef set<LL> SINT;
typedef pair<int,int> PINT;
LL read(){
LL ret=0,f=1;
char x=getchar();
while(!(x>='0' && x<='9')){if(x=='-')f=-1;x=getchar();}
while(x>='0' && x<='9') ret=ret*10+x-'0', x=getchar();
return ret*f;
}
VINT v[10010]; //factor
VINT vec[10010]; // get the position of every number;
//get everynumber's factor
void init() {
for (int i = 1;i < 10010;i++) {
for (int j = 1;j * j <= i;j++) {
if (i % j == 0) {
v[i].push_back(j);
if (j * j != i) v[i].push_back(i / j);
}
}
}
}
int main () {
// FR;
int N;
init();
int a[100010];
while (~scanf("%d",&N)) {
REP(i,10010) {
vec[i].clear();
}
REP(i,N) {
scanf("%d",&a[i]);
vec[a[i]].push_back(i); // get the position
}
int ans = 0;
REP(i,N) {
int mini = N;
int maxn = -1;
REP(j,v[a[i]].size()) {
int factor = v[a[i]][j];
REP(k,vec[factor].size()) {
int pos = vec[factor][k];
if (pos < i) {
maxn = max(maxn,pos);
}
else if (pos > i) {
mini = min(mini,pos);
}
}
}
ans += (mini - i) * (i - maxn);
ans = ans % (int)(1e9 + 7);
}
cout << ans << endl;
}
}
第二题 hdu5289 Assignment
// ST算法 http://blog.csdn.net/insistgogo/article/details/9929103 （ST算法讲解）
#include <map> #include <set> #include <list> #include <stack> #include <queue> #include <cmath> #include <ctime> #include <cstdio> #include <vector> #include <sstream> #include <cstdlib> #include <complex> #include <cstring> #include <iostream> #include <algorithm> #define REP(i,N) for (int i = 0;i < (N);i++) #define REP_1(i,N) for (int i = 1;i < (N);i++) #define REP_2(i,be,en) for (int i = (be);i < (en);i++) #define DWN(i,N) for (int i = (N);i >= 0;i--) #define DWN_1(i,N) for (int i = (N);i >= 1;i--) #define DWN_2(i,en,be) for (int i = (en);i >= (be);i--) #define FR(N) freopen((N),"r",stdin) #define FW(N) freopen((N),"w",stdout) #define MAXN 100000 + 5 #define GETS(ch) fgets((ch),MAXN,stdin) #define INF 0x3f3f3f3f #define MOD 1000000007 using namespace std; typedef long long LL; typedef LL ll; typedef map<int,LL> MINT; typedef map<char,int> MCH; typedef map<string,int> MSTR; typedef vector<int> VINT; typedef set<LL> SINT; typedef pair<int,int> PINT; LL read(){ LL ret=0,f=1; char x=getchar(); while(!(x>='0' && x<='9')){if(x=='-')f=-1;x=getchar();} while(x>='0' && x<='9') ret=ret*10+x-'0', x=getchar(); return ret*f; } int A[MAXN]; int MIN[MAXN][30], MAX[MAXN][30]; void RMQ_init(int N) { REP(i,N) MAX[i][0] = MIN[i][0] = A[i]; for(int j = 1; (1 << j) <= N; j++) { for(int i = 0; i + (1 << j) - 1 < N; i++) { MAX[i][j] = max(MAX[i][j - 1], MAX[i + (1 << (j - 1))][j - 1]); MIN[i][j] = min(MIN[i][j - 1], MIN[i + (1 << (j - 1))][j - 1]); } } } int RMQ_min(int L, int R) { int k = 0; while((1 << (k + 1)) <= R - L + 1) k++; return min(MIN[L][k], MIN[R - (1 << k) + 1][k]); } int RMQ_max(int L, int R) { int k = 0; while((1 << (k + 1)) <= R - L + 1) k++; return max(MAX[L][k], MAX[R - (1 << k) + 1][k]); } int main() { int T; // freopen("1.in","r",stdin); scanf("%d", &T); while(T--) { int N, k; scanf("%d%d", &N, &k); REP(i,N) scanf("%d", &A[i]); RMQ_init(N); LL ans = 0; int pos = 0; REP(i,N) { int low = i, right = N - 1; while(low <= right) { int mid = (low + right) >> 1; if(RMQ_max(i, mid) - RMQ_min(i, mid) < k){ pos = mid; low = mid + 1; } else { right = mid - 1; } } ans += pos - i + 1; } cout << ans << endl; } return 0; }
// 单调队列
#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <vector>
#include <sstream>
#include <cstdlib>
#include <complex>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <deque>
#define REP(i,N) for (int i = 0;i < (N);i++)
#define REP_1(i,N) for (int i = 1;i < (N);i++)
#define REP_2(i,be,en) for (int i = (be);i < (en);i++)
#define DWN(i,N) for (int i = (N);i >= 0;i--)
#define DWN_1(i,N) for (int i = (N);i >= 1;i--)
#define DWN_2(i,en,be) for (int i = (en);i >= (be);i--)
#define FR(N) freopen((N),"r",stdin)
#define FW(N) freopen((N),"w",stdout)
#define MAXN 2010
#define GETS(ch) fgets((ch),MAXN,stdin)
#define INF 0x3f3f3f3f
#define MOD 1000000007
using namespace std;
typedef long long LL;
typedef LL ll;
typedef map<int,LL> MINT;
typedef map<char,int> MCH;
typedef map<string,int> MSTR;
typedef vector<int> VINT;
typedef set<LL> SINT;
typedef pair<int,int> PINT;
LL read(){
LL ret=0,f=1;
char x=getchar();
while(!(x>='0' && x<='9')){if(x=='-')f=-1;x=getchar();}
while(x>='0' && x<='9') ret=ret*10+x-'0', x=getchar();
return ret*f;
}
// DQ1 increase DQ2 decrease
deque<int> DQ1,DQ2;
int a[100010];
int main () {
// FR("1.in");
int T;
cin >> T;
while (T--) {
int N,K;
DQ1.clear();
DQ2.clear();
scanf("%d %d",&N,&K);
REP(i,N) scanf("%d",&a[i]);
ll ans = 0;
for (int i = 0,j = 0;i < N;i++) {
while (!DQ1.empty() && a[DQ1.back()] > a[i]) DQ1.pop_back();
while (!DQ2.empty() && a[DQ2.back()] < a[i]) DQ2.pop_back();
DQ1.push_back(i);
DQ2.push_back(i);
while (a[DQ2.front()] - a[DQ1.front()] >= K) {
if (j == DQ2.front())
DQ2.pop_front();
if (j == DQ1.front())
DQ1.pop_front();
j++;
}
ans += i - j + 1;
}
cout << ans << endl;
}
}
// 树状数组
#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <vector>
#include <sstream>
#include <cstdlib>
#include <complex>
#include <climits>
#include <cstring>
#include <iostream>
#include <algorithm>
#define REP(i,N) for (int i = 0;i < (N);i++)
#define REP_1(i,N) for (int i = 1;i < (N);i++)
#define REP_2(i,be,en) for (int i = (be);i < (en);i++)
#define DWN(i,N) for (int i = (N);i >= 0;i--)
#define DWN_1(i,N) for (int i = (N);i >= 1;i--)
#define DWN_2(i,en,be) for (int i = (en);i >= (be);i--)
#define FR(N) freopen((N),"r",stdin)
#define FW(N) freopen((N),"w",stdout)
#define MAXN 2010
#define GETS(ch) fgets((ch),MAXN,stdin)
#define INF 0x3f3f3f3f
#define MOD 1000000007
using namespace std;
typedef long long LL;
typedef LL ll;
typedef map<int,LL> MINT;
typedef map<char,int> MCH;
typedef map<string,int> MSTR;
typedef vector<int> VINT;
typedef set<LL> SINT;
typedef pair<int,int> PINT;
LL read(){
LL ret=0,f=1;
char x=getchar();
while(!(x>='0' && x<='9')){if(x=='-')f=-1;x=getchar();}
while(x>='0' && x<='9') ret=ret*10+x-'0', x=getchar();
return ret*f;
}
int a[100010];
int N,K;
int maxn[100010];
int mini[100010];
int lowbit(int i) {
return i & (-i);
}
void add(int i,int value) {
while (i <= N) {
maxn[i] = max(maxn[i],value);
mini[i] = min(mini[i],value);
i += lowbit(i);
}
}
void init() {
REP_1(i,N + 1) maxn[i] = INT_MIN;
REP_1(i,N + 1) mini[i] = INT_MAX;
}
int query_min(int i) {
int MIN = INT_MAX;
while (i > 0) {
MIN = min(MIN,mini[i]);
i -= lowbit(i);
}
return MIN;
}
int query_max(int i) {
int MAX = INT_MIN;
while (i > 0) {
MAX = max(MAX,maxn[i]);
i -= lowbit(i);
}
return MAX;
}
int main () {
// FR("1.in");
int T;
cin >> T;
while (T--) {
scanf("%d %d",&N,&K);
init();
REP_1(i,N + 1) scanf("%d",&a[i]);
ll ans = 0;
DWN_1(i,N) {
add(i,a[i]);
int low = i;
int hight = N;
int pos = 0;
while (low <= hight) {
int mid = (hight + low) >> 1;
if (query_max(mid) - query_min(mid) < K) {
pos = mid;
low = mid + 1;
}
else {
hight = mid - 1;
}
}
ans += pos - i + 1;
}
cout << ans << endl;
}
}
第七题 hdu5293 Tricks Device
#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <vector>
#include <sstream>
#include <cstdlib>
#include <complex>
#include <climits>
#include <cstring>
#include <iostream>
#include <algorithm>
#define REP(i,N) for (int i = 0;i < (N);i++)
#define REP_1(i,N) for (int i = 1;i < (N);i++)
#define REP_2(i,be,en) for (int i = (be);i < (en);i++)
#define DWN(i,N) for (int i = (N);i >= 0;i--)
#define DWN_1(i,N) for (int i = (N);i >= 1;i--)
#define DWN_2(i,en,be) for (int i = (en);i >= (be);i--)
#define FR(N) freopen((N),"r",stdin)
#define FW(N) freopen((N),"w",stdout)
#define MAXN 2010
#define GETS(ch) fgets((ch),MAXN,stdin)
#define INF 0x3f3f3f3f
#define MOD 1000000007
using namespace std;
typedef long long LL;
typedef LL ll;
typedef map<int,LL> MINT;
typedef map<char,int> MCH;
typedef map<string,int> MSTR;
typedef vector<int> VINT;
typedef set<LL> SINT;
typedef pair<int,int> PINT;
LL read(){
LL ret=0,f=1;
char x=getchar();
while(!(x>='0' && x<='9')){if(x=='-')f=-1;x=getchar();}
while(x>='0' && x<='9') ret=ret*10+x-'0', x=getchar();
return ret*f;
}
class EdmondsKarp {
private:
struct Edge{
int from,to,flow;
Edge(int u,int v,int f):from(u),to(v),flow(f){}
};
public:
int n;
vector<Edge> edges;
VINT G[MAXN];
int a[MAXN];
int p[MAXN];
int s,t;
int used[MAXN];
void init(int n,int s,int t) {
this -> n = n;
this -> s = s;
this -> t = t;
REP(i,n) G[i].clear();
edges.clear();
}
void AddEdge(int from,int to,int flow) {
edges.push_back(Edge(from,to,flow));
edges.push_back(Edge(to,from,0));
G[from].push_back(edges.size() - 2);
G[to].push_back(edges.size() - 1);
}
int BFS_Maxflow(int s,int t) {
int flow = 0;
while (1) {
memset(a,0,sizeof(a));
queue<int> Q;
Q.push(s);
a[s] = INF;
while(!Q.empty()) {
int x = Q.front();Q.pop();
REP(i,G[x].size()) {
Edge& e = edges[G[x][i]];
if (!a[e.to] && e.flow > 0) {
p[e.to] = G[x][i];
a[e.to] = min(a[x],e.flow);
Q.push(e.to);
}
}
if (a[t]) break;
}
if (!a[t]) break;
for (int i = t;i != s;i = edges[p[i]].from) {
edges[p[i]].flow -= a[t];
edges[p[i]^1].flow += a[t];
}
flow += a[t];
}
return flow;
}
int dfs(int s,int t,int f) {
if (s == t) return f;
used[s] = 1;
REP(i,G[s].size()) {
Edge& e = edges[G[s][i]];
if (!used[e.to] && e.flow > 0) {
int d = dfs(e.to,t,min(f,e.flow));
if (d > 0) {
e.flow -= d;
edges[G[s][i]^1].flow += d;
return d;
}
}
}
return 0;
}
int DFS_Maxflow(int s,int t) {
int flow = 0;
while (1) {
memset(used,0,sizeof(used));
int f = dfs(s,t,INF);
if (f == 0) return flow;
flow += f;
}
return flow;
}
}E;
class Dijkstra {
private:
struct Edge{
int from,to,dist;
Edge(int u,int v,int d):from(u),to(v),dist(d) {}
};
int n;
vector<Edge> edges;
VINT G[MAXN];
bool done[MAXN];
int d[MAXN];
VINT path[MAXN];
int number[MAXN];
public:
void init(int n) {
this -> n = n;
edges.clear();
REP(i,n) {
G[i].clear();
number[i] = INF;
path[i].clear();
}
}
void AddEdge(int from,int to,int dist){
edges.push_back(Edge(from,to,dist));
G[from].push_back(edges.size() - 1);
}
void dijkstra(int s) {
priority_queue<PINT,vector<PINT>,greater<PINT> > Q;
REP(i,n) d[i] = INF;
d[s] = 0;
memset(done,false,sizeof(done));
Q.push(make_pair(d[s],s));
while (!Q.empty()) {
PINT P = Q.top();Q.pop();
int u = P.second;
if (done[u]) continue;
done[u] = true;
REP(i,G[u].size()) {
Edge& e = edges[G[u][i]];
if (d[e.to] > d[u] + e.dist) {
d[e.to] = d[u] + e.dist;
path[e.to].clear();
path[e.to].push_back(u);
Q.push(make_pair(d[e.to],e.to));
}
else if (d[e.to] == d[u] + e.dist){
path[e.to].push_back(u);
Q.push(make_pair(d[e.to],e.to));
}
}
}
}
int dfs(int end,int s) {
if (end == s) return 0;
if (number[end] != INF) return number[end];
REP(i,path[end].size()) {
int fa = path[end][i];
E.AddEdge(fa,end,1);
number[end] = min(number[end],dfs(fa,s) + 1);
}
return number[end];
}
}D;
int main () {
// FR("1.in");
int N,M;
while (~scanf("%d %d",&N,&M)) {
D.init(N);
REP(i,M) {
int u,v,c;
scanf("%d %d %d",&u,&v,&c);
u --; v --;
D.AddEdge(u,v,c);
D.AddEdge(v,u,c);
}
D.dijkstra(0);
int dis = D.dfs(N - 1,0);
cout << E.DFS_Maxflow(0,N - 1) << " " << M - dis << endl;
}
}
