PKU 2892 [HDU 1540]

Tunnel Warfare

题型：线段树

描述：n个村子排成一条线，两两相邻，1.摧毁村子c，2.修复上一次摧毁的村子，3.问与村子c直接或间接相通的村子数

思路：对节点的操作，增加lx,rx域，表示区间内最左边被摧毁的村子，和最右边被摧毁的村子，s表示区间内摧毁的村子总数。

重点在count()，统计相连村子数操作。根据父结点的两个孩子结点的 s 值，讨论各种可能情况。

if (t[k0].s == 0) {
    if (t[k0+1].s == 0) cnt = t[k].r-t[k].l+1;
    else if (t[k0+1].lp > a) cnt = t[k0].r-t[k0].l+1+t[k0+1].lp-t[k0+1].l;
    else count(a, k0+1);
}else {
    if (t[k0+1].s == 0 && t[k0].rp < a) 
        cnt = t[k0+1].r-t[k0+1].l+1+t[k0].r-t[k0].rp;    
    else if (a>t[k0].rp && a<t[k0+1].lp) 
        cnt = t[k0].r-t[k0].rp + t[k0+1].lp-t[k0+1].l;
    else {
        if (a <= md) count(a, k0);
        else count(a, k0+1);
    }
}

 

//234MS    2804K
#include <stdio.h>
#include <string.h>
#define NL 65536

struct Seg {
    int l, r;
    int lp, rp, s;
}t[NL*2];
bool dt[50001];//1 没被摧毁 0 被摧毁
int stk[50001];
int cnt;

void build(int l, int r, int k)
{
    t[k].l = l;
    t[k].r = r;
    t[k].s = 0;
    t[k].lp = t[k].rp = 0;
    if (l == r) dt[l] = 1;
    int md = (l+r)>>1, k0 = k<<1;
    if (l < r) {
        build(l, md, k0);
        build(md+1, r, k0+1);
    }
}
//flg = 0 摧毁,flg = 1 恢复
void mody(int a, int k, bool flg)
{
    if (t[k].l == t[k].r) {
        if (flg) t[k].s = 0, dt[a] = 1;
        else {
            t[k].s = 1;
            t[k].lp = t[k].rp = a;
            dt[a] = 0;
        }
        return;
    }
    int md = (t[k].l+t[k].r)>>1, k0 = k<<1;
    if (a <= md) 
        mody(a, k0, flg);
    else if (a > md)
        mody(a, k0+1, flg);
    t[k].s = t[k0].s + t[k0+1].s;
    if (t[k].s) {
        t[k].lp = t[k0].lp;
        t[k].rp = t[k0+1].rp;
        if (t[k0].s == 0) t[k].lp = t[k0+1].lp;
        if (t[k0+1].s == 0) t[k].rp = t[k0].rp;
    }
}

void count(int a, int k)
{
    if (t[k].l == t[k].r) {
        cnt = 1;
        return;
    }
    int md = (t[k].l+t[k].r)>>1, k0 = k<<1;
    if (t[k0].s == 0) {
        if (t[k0+1].s == 0) cnt = t[k].r-t[k].l+1;
        else if (t[k0+1].lp > a) cnt = t[k0].r-t[k0].l+1+t[k0+1].lp-t[k0+1].l;
        else count(a, k0+1);
    }else {
        if (t[k0+1].s == 0 && t[k0].rp < a) 
            cnt = t[k0+1].r-t[k0+1].l+1+t[k0].r-t[k0].rp;    
        else if (a>t[k0].rp && a<t[k0+1].lp) 
            cnt = t[k0].r-t[k0].rp + t[k0+1].lp-t[k0+1].l;
        else {
            if (a <= md) count(a, k0);
            else count(a, k0+1);
        }
    }
}

int main()
{
    int n, m, a, top;
    char c[3];
//    freopen("in.txt", "r", stdin);
    while (scanf("%d%d", &n, &m) != EOF) {
        build(1, n, 1);
        top = 0;
        while (m--) {
            scanf("%s", c);
            if (c[0] == 'D') {
                scanf("%d", &a);
                stk[top++] = a;
                mody(a, 1, 0);
            }else if (c[0] == 'Q') {
                scanf("%d", &a);
                cnt = 0;
                if (dt[a])
                    count(a, 1);
                printf("%d\n", cnt);
            }else {
                a = stk[--top];
                mody(a, 1, 1);
            }
        }
    }
    return 0;
}