feat: 添加P5835题解实现，计算最小碰撞次数

21/100pts
实现了一个算法来计算在特定条件下粒子碰撞的最小次数。主要包含以下功能：
1. 读取输入数据并初始化粒子信息
2. 根据粒子运动方向计算碰撞时间
3. 使用归并排序统计逆序对数作为碰撞次数

src/10/22/P11670.cpp

@@ -1,11 +1,53 @@
+#include <algorithm>
 #include <cstdint>
+#include <cstdio>
 #include <iostream>
 #include <istream>
+#include <vector>
 using ll = int64_t;
 
-
+ll n,ans;
+std::vector<std::vector<ll>> apos,bpos;
+std::vector<ll> a,b;
 
 int main(){
     std::iostream::sync_with_stdio(false);
     std::cin.tie(nullptr);
+
+    std::cin>>n;
+
+    a.resize(n+1);
+    b.resize(n+1);
+    apos.resize(n+1);
+    bpos.resize(n+1);
+
+    for(ll i=1;i<=n;i++){
+        std::cin>>a[i];
+        apos[a[i]].push_back(std::min(i,n-i+1));
+    }
+    for(ll i=1;i<=n;i++){
+        std::cin>>b[i];
+        bpos[b[i]].push_back(std::min(i,n-i+1));
+    }
+    for(ll i=1;i<=n;i++){
+        if(a[i]==b[i]){
+            ll N=i-1;
+            ans+=N*(N+1)/2;
+            N=n-i;
+            ans+=N*(N+1)/2;
+        }
+        std::sort(apos[i].begin(),apos[i].end());
+        std::sort(bpos[i].begin(),bpos[i].end());
+
+        ll pa=0,pb=0;
+        while(pa<apos[i].size() && pb<bpos[i].size()){
+            if(apos[i][pa]<bpos[i][pb]){
+                ans+=apos[i][pa++]*(bpos[i].size()-pb);
+            }else{
+                ans+=bpos[i][pb++]*(apos[i].size()-pa);
+            }
+        }
+    }
+
+    std::cout<<ans<<"\n";
 }

src/10/22/P5835.cpp

@@ -0,0 +1,102 @@
+#include <algorithm>
+#include <cstdint>
+#include <cstdio>
+#include <iostream>
+#include <istream>
+#include <map>
+#include <unordered_set>
+#include <vector>
+using ll = int64_t;
+
+const ll maxn = 5e4+5;
+ll n,l,allweight;
+struct P{
+    ll w,x,d;
+    inline bool operator<(const P&other)const{
+        return x<other.x;
+    }
+}c[maxn];
+struct D{
+    ll d,idx;
+    inline bool operator<(const D&other)const{
+        return d<other.d;
+    }
+}dis[maxn];
+std::vector<ll> after,rafter,tmp;
+std::unordered_set<ll> vis;
+
+static inline ll merge(std::vector<ll>&num,std::vector<ll>&tmp,ll l,ll mid,ll r){
+    ll cnt=0;
+    ll i=l,j=mid+1,k=l;
+    while(i<=mid&&j<=r){
+        if(num[i]<=num[j]){
+            tmp[k++]=num[i++];
+        }else{
+            cnt+= mid-i+1;
+            tmp[k++]=num[j++];
+        }
+    }
+    while(i<=mid)tmp[k++]=num[i++];
+    while(j<=r)tmp[k++]=num[j++];
+    for(ll i=l;i<=r;i++){
+        num[i]=tmp[i];
+    }
+    return cnt;
+}
+
+static inline ll mergesort(std::vector<ll>&num,std::vector<ll>&tmp,ll l,ll r){
+    ll cnt=0;
+    if(l<r){
+        ll mid=(l+r)/2;
+        cnt+=mergesort(num, tmp, l, mid);
+        cnt+=mergesort(num, tmp, mid+1, r);
+        cnt+=merge(num,tmp,l,mid,r);
+    }
+    return cnt;
+}
+
+int main(){
+    std::iostream::sync_with_stdio(false);
+    std::cin.tie(nullptr);
+
+    std::cin>>n>>l;
+    after.reserve(n+1);
+    tmp.resize(n+1);
+    for(ll i=1;i<=n;i++){
+        std::cin>>c[i].w>>c[i].x>>c[i].d;
+        allweight+=c[i].w;
+    }
+    std::sort(c+1,c+1+n);
+    for(ll i=1;i<=n;i++){
+        dis[i].idx=i;
+        if(c[i].d==1){
+            dis[i].d=l-c[i].x;
+        }else{
+            dis[i].d=c[i].x;
+        }
+    }
+    std::sort(dis+1,dis+1+n);
+    ll outweight=0;
+    ll pout=0;
+    for(ll i=1;i<=n;i++){
+        outweight+=c[dis[i].idx].w;
+        // printf("%lld is out\n",c[dis[i].idx].x);
+        pout=dis[i].d;
+        vis.emplace(dis[i].idx);
+        if(outweight*2>=allweight){
+            break;
+        }
+    }
+    // printf("lnum=%lld, rnum=%lld\n",lnum,rnum);
+    // std::cout<<"after:\n";
+    // for(ll i:after){
+    //     std::cout<<i<<" ,";
+    // }
+    // std::cout<<"\n";
+    after.push_back(0);
+    for(ll i=1;i<=n;i++){
+        after.push_back(c[i].x+c[i].d*pout);
+    }
+    
+    std::cout<<mergesort(after, tmp, 1, n)<<"\n";
+}