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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#65059 | #4622. Mario Party | qinjianbin | AC ✓ | 5286ms | 95252kb | C++ | 6.0kb | 2022-11-26 21:55:31 | 2022-11-26 21:55:31 |
Judging History
answer
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<cctype>
#include<ctime>
#include<queue>
#include<stack>
#include<deque>
#include<map>
#include<unordered_map>
#include<set>
#include<unordered_set>
#include<bitset>
#include<functional>
using namespace std;
//STL库常用(vector, map, set, pair)
#define pii pair<int,int>
#define pdd pair<double,double>
#define F first
#define S second
//常量
#define PI (acos(-1.0))
#define INF 0x3f3f3f3f
//必加
#ifdef TanJI
#include<cassert>
#define dbg(...) do {cerr << #__VA_ARGS__ << " -> "; err(__VA_ARGS__); } while(0)
void err() {cerr << endl;}
template<template<typename...> class T, typename t, typename... A> void err(T<t> a, A... x) {for (auto v: a) cerr << v << ' '; cerr << ", "; err(x...);}
template<typename T, typename... A> void err(T a, A... x) {cerr << a << ' '; err(x...);}
template<typename T> void err(T *a, int len) {for (int i = 0; i < len; i++) cerr << a[i] << ' '; err();}
#else
#define dbg(...)
#define assert(...)
#endif
typedef unsigned long long ull;
typedef long long ll;
typedef double db;
// 快读快输
template<typename T> inline T read() {T x=0; bool f=0; char c=getchar();while(!isdigit(c)){if(c =='-')f= 1; c=getchar();}while(isdigit(c)){x=(x<<3)+(x<<1)+(c^48);c=getchar();}return f?~x+1:x;}
template<typename T> inline void print(T k){ int num = 0,ch[20]; if(k == 0){ putchar('0'); return ; } (k<0)&&(putchar('-'),k = -k); while(k>0) ch[++num] = k%10, k /= 10; while(num) putchar(ch[num--]+48); }
//图论
const int GRAPHN = 1, GRAPHM = 1;
int h[GRAPHN], e[GRAPHM], ne[GRAPHM], wei[GRAPHM], idx;
void AddEdge(int a, int b) {e[idx] = b, ne[idx] = h[a], h[a] = idx++;}
void AddEdge(int a, int b, int c) {e[idx] = b, ne[idx] = h[a], wei[idx] = c, h[a] = idx++;}
void AddEdge(int h[], int a, int b) {e[idx] = b, ne[idx] = h[a], h[a] = idx++;}
void AddEdge(int h[], int a, int b, int c) {e[idx] = b, ne[idx] = h[a], wei[idx] = c, h[a] = idx++;}
// 数学公式
template<typename T> inline T gcd(T a, T b){ return b==0 ? a : gcd(b,a%b); }
template<typename T> inline T lowbit(T x){ return x&(-x); }
template<typename T> inline bool mishu(T x){ return x>0?(x&(x-1))==0:false; }
template<typename T1,typename T2, typename T3> inline ll q_mul(T1 a,T2 b,T3 p){ ll w = 0; while(b){ if(b&1) w = (w+a)%p; b>>=1; a = (a+a)%p; } return w; }
template<typename T,typename T2> inline ll f_mul(T a,T b,T2 p){ return (a*b - (ll)((long double)a/p*b)*p+p)%p; }
template<typename T1,typename T2, typename T3> inline ll q_pow(T1 a,T2 b,T3 p){ ll w = 1; while(b){ if(b&1) w = (w*a)%p; b>>=1; a = (a*a)%p;} return w; }
template<typename T1,typename T2, typename T3> inline ll s_pow(T1 a,T2 b,T3 p){ ll w = 1; while(b){ if(b&1) w = q_mul(w,a,p); b>>=1; a = q_mul(a,a,p);} return w; }
template<typename T> inline ll ex_gcd(T a, T b, T& x, T& y){ if(b == 0){ x = 1, y = 0; return (ll)a; } ll r = ex_gcd(b,a%b,y,x); y -= a/b*x; return r;/*gcd*/ }
template<typename T1,typename T2> inline ll com(T1 m, T2 n) { int k = 1;ll ans = 1; while(k <= n){ ans=((m-k+1)*ans)/k;k++;} return ans; }
template<typename T> inline bool isprime(T n){ if(n <= 3) return n>1; if(n%6 != 1 && n%6 != 5) return 0; T n_s = floor(sqrt((db)(n))); for(int i = 5; i <= n_s; i += 6){ if(n%i == 0 || n%(i+2) == 0) return 0; } return 1; }
/* ----------------------------------------------------------------------------------------------------------------------------------------------------------------- */
const int MAXN = 1e6 + 10;
int T, n, q;
int a[MAXN];
int fa[MAXN];
int Find(int u) {
int a = u;
while (u != fa[u])
u = fa[u];
while (a != fa[a]) {
int b = a;
a = fa[a];
fa[b] = u;
}
return u;
}
int pos[MAXN], invp[MAXN * 3]; // 每个询问指向的值,值反函数对应哪些询问并查集的头节点(用于合并询问)
void Merge(int val1, int val2) {
if (invp[val1] == -1) return;
if (invp[val2] == -1) {
invp[val2] = invp[val1];
invp[val1] = -1;
pos[invp[val2]] = val2;
return;
}
fa[invp[val1]] = invp[val2];
invp[val1] = -1;
}
map<int, vector<int> > add, del;
int x[MAXN], ans[MAXN];
void init() {
for (int i = 1; i <= q; i++)
fa[i] = i;
for (int i = 1; i <= q; i++)
pos[i] = -1;
for (int i = 0; i <= 3e6; i++)
invp[i] = -1;
add.clear();
del.clear();
}
int main() {
#ifdef TanJI
clock_t c1 = clock();
freopen("D:\\Cpp\\1.in", "r", stdin);
freopen("D:\\Cpp\\1.out", "w", stdout);
#endif
//--------------------------------------------
scanf("%d", &T);
while (T--) {
scanf("%d%d", &n, &q);
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
init();
int l, r;
for (int i = 1; i <= q; i++) {
scanf("%d%d%d", &l, &r, &x[i]);
add[l].push_back(i);
del[r].push_back(i);
}
int start = 1e6;
for (int i = 1; i <= n; i++) {
if (a[i] < 0)
for (int j = start; j < start - a[i]; j++)
Merge(j, j - a[i]);
start -= a[i];
// dbg(start);
if (add.count(i))
for (auto u : add[i]) {
int now = x[u] + start;
if (invp[now] == -1) {
pos[u] = now;
invp[now] = u;
// dbg(now);
} else {
// merege
fa[u] = invp[now];
}
}
if (del.count(i))
for (auto u : del[i]) {
ans[u] = pos[Find(u)] - start;
// dbg(u, Find(u), pos[Find(u)], start);
}
}
for (int i = 1; i <= q; i++)
printf("%d\n", ans[i]);
}
//--------------------------------------------
#ifdef TanJI
cerr << "Time Used:" << clock() - c1 << "ms" << endl;
#endif
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 5286ms
memory: 95252kb
input:
4 500000 500000 2 -1 2 -1 1 -1 -4 -2 0 -3 4 1 -3 3 5 1 -3 -2 0 3 -2 0 -1 0 0 -3 -2 0 0 1 -2 -3 -4 2 5 0 -1 -2 2 2 -2 -1 2 0 0 -2 3 -2 -4 -3 -3 -1 0 2 1 0 2 -1 -3 0 -2 3 2 -2 1 4 -1 -1 -2 0 2 -2 1 -4 -5 -1 -2 -3 -2 1 4 2 -3 0 1 -1 2 2 0 5 -3 0 2 -1 5 -3 5 -4 -4 0 2 2 1 -1 -6 0 2 -3 2 3 0 0 -2 3 4 2 1...
output:
1363 427 212 818 736 132 561 623 81 800 12 971 218 28 278 737 581 363 25 231 216 642 684 283 1033 982 469 106 156 197 992 384 230 112 226 38 31 561 637 43 623 246 397 1640 1252 1354 4 1338 168 266 364 68 10 435 1391 229 106 0 1114 1416 201 416 206 113 474 190 612 206 109 670 34 0 804 604 376 164 251...
result:
ok 2000000 lines