Luogu2018 消息传递（树形DP）

最新推荐文章于 2020-11-10 16:12:50 发布

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最新推荐文章于 2020-11-10 16:12:50 发布

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/bingoyes/p/11255823.html

本文介绍了一个使用贪心算法解决子树节点计数问题的C++实现。通过预处理将节点分配到不同块中，实现了快速查询和更新操作。文章详细展示了代码结构，包括自定义输入输出流、数据结构定义、主要算法流程和复杂度分析。

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贪心优先子树较多者。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int  a = (b); a <= (c); ++ a)
#define nR(a,b,c) for(register int  a = (b); a >= (c); -- a)
#define Max(a,b) ((a) > (b) ? (a) : (b))
#define Min(a,b) ((a) < (b) ? (a) : (b))
#define Fill(a,b) memset(a, b, sizeof(a))
#define Abs(a) ((a) < 0 ? -(a) : (a))
#define Swap(a,b) a^=b^=a^=b
#define ll long long

#define ON_DEBUG

#ifdef ON_DEBUG

#define D_e_Line printf("\n\n----------\n\n")
#define D_e(x)  cout << #x << " = " << x << endl
#define Pause() system("pause")
#define FileOpen() freopen("in.txt","r",stdin);

#else

#define D_e_Line ;
#define D_e(x)  ;
#define Pause() ;
#define FileOpen() ;

#endif

struct ios{
    template<typename ATP>ios& operator >> (ATP &x){
        x = 0; int f = 1; char c;
        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-')  f = -1;
        while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
        x*= f;
        return *this;
    }
}io;
using namespace std;

const int N = 50007;

int n;

int vis[N];
int sta[N], top;
int block[N], blockSize;
int ans[N], peace[N];
inline void Destroy(int x){
    vis[x] = true;
    --ans[block[x]];
}
inline void Rescue(int x){
    vis[x] = false;
    ++ans[block[x]];
}
inline int Query(int x){
    if(vis[x] == true) return 0;
    int sum = 0;
    int flagLeft = 0, flagRight = 0;
    int maxx = Min(block[x] * blockSize, n);
    R(i,x + 1,maxx){
        if(vis[i] == true){
            flagRight = 1;
            break;
        }
        ++sum;
    }
    nR(i,x - 1, (block[x] - 1) * blockSize + 1){
        if(vis[i] == true){
            flagLeft = 1;
            break;
        }
        ++sum;
    }
    if(flagLeft == 1 && flagRight == 1) return sum + 1;
    
    int flag = 0;
    if(flagRight == 0){
        R(i,block[x] + 1, block[n]){
            if(flag) break;
            if(ans[i] == peace[i]){
                sum += ans[i];
                continue;
            }
            flag = 1;
            int maxx = Min(i * blockSize, n);
            R(j,(i - 1) * blockSize + 1, maxx){
                if(vis[j] == true) break;
                ++sum;
            }
        }       
    }

    if(flagLeft == 0){
        flag = 0;
        nR(i,block[x] - 1, 1){
            if(flag) break;
            if(ans[i] == peace[i]){
                sum += ans[i];
                continue;
            }
            flag = 1;
            nR(j,i * blockSize, (i - 1) * blockSize + 1){
                if(vis[j] == true){
                    break;
                }
                ++sum;
            }
        }       
    }

    return sum + 1;
}

int main(){
//FileOpen();
    int m;
    io >> n >> m;
    blockSize = sqrt(n) + 1;
    R(i,1,n){
        block[i] = (i - 1) / blockSize + 1;
        ++ans[block[i]];
    }
    R(i,1,blockSize) peace[i] = ans[i];
    while(m--){
        char ch = getchar();
        while(ch != 'D' && ch != 'R' && ch != 'Q') ch = getchar();
        if(ch == 'D'){
            int x;
            io >> x;
            Destroy(x);
            sta[++top] = x;
        }
        else if (ch == 'R'){
            if(!top) continue;
            Rescue(sta[top]);
            --top;
        }
        else{
            int x;
            io >> x;
            printf("%d\n", Query(x));
        }
    }
    
    return 0;
}
/*
$1 belong to 1
$2 belong to 1
$3 belong to 2
$4 belong to 2
$5 belong to 3
$6 belong to 3
$7 belong to 4
*/