HDU 1540 Tunnel Warfare

线段树，区间合并。n个村子连成一条线，鬼子炸，八路修，问x村左右共联通多少村子。三种操作分别对应炸修和问。模拟个栈来判断该修哪个，栈空判断一下。在开个数组记录一下目前各村状态，炸了就别再炸了，修了就别再修了。

建树时，三个数组的初始值都是当前区间长度。更新时，判断目标p在左边还是右边。询问时，先看p在哪边，假设p在左子树，并且在左子树的rsum内，也就是说可以一直连到右子树，那么就加上lsum[rt << 1 | 1]。反之亦然

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<vector>#include<queue>#include<cmath>///LOOP#define REP(i, n) for(int i = 0; i < n; i++)#define FF(i, a, b) for(int i = a; i < b; i++)#define FFF(i, a, b) for(int i = a; i <= b; i++)#define FD(i, a, b) for(int i = a - 1; i >= b; i--)#define FDD(i, a, b) for(int i = a; i >= b; i--)///INPUT#define RI(n) scanf("%d", &n)#define RII(n, m) scanf("%d%d", &n, &m)#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)#define RFI(n) scanf("%lf", &n)#define RFII(n, m) scanf("%lf%lf", &n, &m)#define RFIII(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)#define RFIV(n, m, k, p) scanf("%lf%lf%lf%lf", &n, &m, &k, &p)#define RS(s) scanf("%s", s)///OUTPUT#define PN printf("\n")#define PI(n) printf("%d\n", n)#define PIS(n) printf("%d ", n)#define PS(s) printf("%s\n", s)#define PSS(s) printf("%s ", n)#define PC(n) printf("Case %d: ", n)///OTHER#define PB(x) push_back(x)#define CLR(a, b) memset(a, b, sizeof(a))#define CPY(a, b) memcpy(a, b, sizeof(b))#define display(A, n, m) {REP(i, n){REP(j, m)PIS(A[i][j]);PN;}}#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1using namespace std;typedef long long LL;typedef pair<int, int> P;const int MOD = 9901;const int INFI = 1e9 * 2;const LL LINFI = 1e17;const double eps = 1e-6;const double pi = acos(-1.0);const int N = 55555;const int M = 22;const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};int s[N], lsum[N << 2], rsum[N << 2], msum[N << 2];bool d[N];void pushup(int rt, int m){    lsum[rt] = lsum[rt << 1];    rsum[rt] = rsum[rt << 1 | 1];    if(lsum[rt] == (m - (m >> 1)))lsum[rt] += lsum[rt << 1 | 1];    if(rsum[rt] == (m >> 1))rsum[rt] += rsum[rt << 1];    msum[rt] = max(lsum[rt << 1 | 1] + rsum[rt << 1], max(msum[rt << 1], msum[rt << 1 | 1]));}void build(int l, int r, int rt){    lsum[rt] = rsum[rt] = msum[rt] = r - l + 1;    if(l == r)return;    int m = (l + r) >> 1;    build(lson);    build(rson);}void update(int p, int v, int l, int r, int rt){    if(l == r)    {        lsum[rt] = rsum[rt] = msum[rt] = v;        return;    }    int m = (l + r) >> 1;    if(p <= m)update(p, v, lson);    else update(p, v, rson);    pushup(rt, r - l + 1);}int query(int p, int l, int r, int rt){    if(l == r || msum[rt] == 0 || msum[rt] == r - l + 1)return msum[rt];    int m = (l + r) >> 1;    if(p <= m)    {        if(p + rsum[rt << 1] > m)return query(p, lson) + lsum[rt << 1 | 1];        else return query(p, lson);    }    else    {        if(p - lsum[rt << 1 | 1] <= m)return query(p, rson) + rsum[rt << 1];        else return query(p, rson);    }}int main(){    //freopen("input.txt", "r", stdin);    char op[5];    int n, m, a, num;    while(RII(n, m) != EOF)    {        num = 0;        build(1, n, 1);        REP(i, n + 1)d[i] = 1;        REP(i, m)        {            RS(op);            if(op[0] == 'D')            {                RI(a);                if(d[a])                {                    d[a] = 0;                    update(a, 0, 1, n, 1);                }                s[num++] = a;            }            else if(op[0] == 'R')            {                if(!num)continue;                a = s[--num];                if(!d[a])                {                    d[a] = 1;                    update(a, 1, 1, n, 1);                }            }            else            {                RI(a);                PI(query(a, 1, n, 1));            }        }    }    return 0;}