题意：蜗牛爬树问题；值得一提的是在第n天如果恰好在天黑时爬到END，则恰好整除，不用再+1；
day = (End - Begin - day0)/(12*(up-down))+1;
#include <iostream> #include <algorithm> #include <stdlib.h> #include <time.h> #include <cmath> #include <cstdio> #include <string> #include <cstring> #include <vector> #include <queue> #include <stack> #include <set> #define c_false ios_base::sync_with_stdio(false); cin.tie(0) #define INF 0x3f3f3f3f #define INFL 0x3f3f3f3f3f3f3f3f #define zero_(x,y) memset(x , y , sizeof(x)) #define zero(x) memset(x , 0 , sizeof(x)) #define MAX(x) memset(x , 0x3f ,sizeof(x)) #define swa(x,y) {LL s;s=x;x=y;y=s;} using namespace std ; #define N 50005 const double PI = acos(-1.0); typedef long long LL ; int main(){ //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); //ios_base::sync_with_stdio(false); cin.tie(0); int Begin, End, up, down; scanf("%d%d%d%d", &Begin, &End, &up, &down); int day0 = 8*up; int day; if(up-down <= 0){ if(End - Begin - day0 <=0) day = 0; else day = -1; }else{ if(End - Begin - day0 <=0) day = 0; else { if((End - Begin - day0)%(12*(up-down)) == 0) day = (End - Begin - day0)/(12*(up-down)); else day = (End - Begin - day0)/(12*(up-down))+1; } } cout<<day; return 0; }
B:z-sort
题意：排序题；
#include <iostream> #include <algorithm> #include <stdlib.h> #include <time.h> #include <cmath> #include <cstdio> #include <string> #include <cstring> #include <vector> #include <queue> #include <stack> #include <set> #define c_false ios_base::sync_with_stdio(false); cin.tie(0) #define INF 0x3f3f3f3f #define INFL 0x3f3f3f3f3f3f3f3f #define zero_(x,y) memset(x , y , sizeof(x)) #define zero(x) memset(x , 0 , sizeof(x)) #define MAX(x) memset(x , 0x3f ,sizeof(x)) #define swa(x,y) {LL s;s=x;x=y;y=s;} using namespace std ; #define N 10005 const double PI = acos(-1.0); typedef long long LL ; int main(){ //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); //ios_base::sync_with_stdio(false); cin.tie(0); int n, a[N], t[N]; scanf("%d", &n); for(int i = 0; i < n; i++){ scanf("%d" ,&a[i]); } sort(a, a+n); int i = 0, j = 0; for(; i < n; i+=2, j++) t[i] = a[j]; for(i = 1; i < n; i+=2, j++) t[i] = a[j]; for(int k = 0; k < n ; k++) printf("%d ", t[k]); return 0; }
ans=总区间数−非法的
经典离线；
所有区间如下：
(1,1),(2,2),(3,3),(4,4),(5,5)……(n,n)
(1,2),(2,3),(3,4),(4,5)……(n-1,n)
(1,3),(2,4),(3,5) ……(n-2,n)
(1,4),(2,5)……(n-3,n)
(1,5)……(n-4,n)
(1,n)
若给出非法对(2,4)
则非法区间是(1,4)与(2,4)以及其下方的所有区间；总数为:(n - 4+1)*(2);
要想在计算区间是不出现重复，则对于所有right值相同的非法对，只取left值最大的；
然后再对right值进行枚举：for：1->n;
ans += 1LL*(z[i]-pos) * (n - i +1);
pos = z[i];
也可以对left枚举；（此时的排序方法与上述不同）
时间复杂度O(nlogn)
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <ctime>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#define c_false ios_base::sync_with_stdio(false); cin.tie(0)
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
#define zero_(x,y) memset(x , y , sizeof(x))
#define zero(x) memset(x , 0 , sizeof(x))
#define MAX(x) memset(x , 0x3f ,sizeof(x))
#define swa(x,y) {LL s;s=x;x=y;y=s;}
using namespace std ;
#define N 300005
const double PI = acos(-1.0);
typedef long long LL ;
int a[N], z[N], m, n;
int main(){
//reopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
//ios_base::sync_with_stdio(false); cin.tie(0);
scanf("%d%d", &n, &m);
zero(a);zero(z);
int x;
for(int i = 1; i <= n ; i++){
scanf("%d", &x);
a[x] = i;
}
int c,v;
for(int i = 1; i <= m; i++){
scanf("%d%d", &c, &v);
int t = max(a[c], a[v]);
z[t] = max(z[t], min(a[c], a[v]));
}
LL ans = 0, pos = 0;
for(int i = 1; i<= n ;i++){
if(z[i] -pos >0){
ans += 1LL*(z[i]-pos) * (n - i +1);
pos = z[i];
}
}
printf("%I64d
", 1LL*n*(n+1)/2-ans);
return 0;
}
思路: 离散化+树状数组
离散化：如果数据是2，10，1000，100000；则存储所需空间是1e5*4；而离散化存储只需要5*4，
数据只需要保证其位置不变即可，将2，10，1000，100000用1，2，3，4代替；
其位置关系，大小关系都不变，但是存储空间变小了；
这样更方便存储，使得使用树状数组成为可能；
剩下的就是树状数组了：按左区间排下序，然后累计右区间数内区间数；（按左区间从大到小累计，否则，嘿嘿）
时间复杂度：O(nlgn);
#include <iostream> #include <algorithm> #include <cstdlib> #include <ctime> #include <cmath> #include <cstdio> #include <string> #include <cstring> #include <vector> #include <queue> #include <stack> #include <set> #define c_false ios_base::sync_with_stdio(false); cin.tie(0) #define INF 0x3f3f3f3f #define INFL 0x3f3f3f3f3f3f3f3f #define zero_(x,y) memset(x , y , sizeof(x)) #define zero(x) memset(x , 0 , sizeof(x)) #define MAX(x) memset(x , 0x3f ,sizeof(x)) #define swa(x,y) {LL s;s=x;x=y;y=s;} using namespace std ; #define N 200005 #define lowbit(k) k&(-k) const double PI = acos(-1.0); typedef long long LL ; struct BIT{int l, r, id;}; int bit[N],ANS[N],n; BIT num[N]; void update(int s, int k){ for(int j = s; j <= n; j +=lowbit(j)) bit[j] += k; } int query(int k){ int ans = 0; for(int i = k; i > 0; i -=lowbit(i)) ans += bit[i]; return ans; } bool compL(BIT a, BIT b){return a.l<b.l;} bool compR(BIT a, BIT b){return a.r<b.r;} int main(){ //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); //ios_base::sync_with_stdio(false); cin.tie(0); scanf("%d",&n); for(int i = 1; i <= n; i++){ scanf("%d%d", &num[i].l, &num[i].r); num[i].id = i; } sort(num+1, num+n+1, compR); //for(int i = 1; i <= n; i++) cout<<num[i].l<<" "<<num[i].r<<" "<<num[i].id<<endl; for(int i = 1; i <= n; i++) num[i].r = i; sort(num+1, num+n+1, compL); //for(int i = 1; i <= n; i++) cout<<num[i].l<<" "<<num[i].r<<" "<<num[i].id<<endl; for(int i = n;i>=1; i--){ ANS[num[i].id] = query(num[i].r); update(num[i].r, 1); } for(int i = 1; i <= n; i++) printf("%d ", ANS[i]); return 0; }
思路：强连通分量
智商太低，无法理解，还是滚去看书去了；
题解连接：
http://www.cnblogs.com/Recoder/p/5323546.html
思路：先做了poj 1852再说；