hdu 1540 Tunnel Warfare(线段树)

题目链接：hdu 1540 Tunnel Warfare

题目大意：有连续的N个城镇，三种操作：

D x：第x城镇被破坏
Q x：插叙第x城镇所在联通块有多少个城镇没有被破坏
R：修复最后一个被破坏的城镇

解题思路：线段树区间合并。每一个城镇看成一个叶子节点，用一个vector记录破坏顺序。对于查询来说，每次仅仅要推断是否在mid−R[lson(u)],mid+L[rson(u)]之间就可以，否则即递归查询左右子树。

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

const int maxn = 50005;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2];
int L[maxn << 2], R[maxn << 2], S[maxn << 2];

inline int length(int u) {
return rc[u] - lc[u] + 1;
}

inline void maintain(int u, int v) {
S[u] = L[u] = R[u] = (v ? length(u) : 0);
}

inline void pushup (int u) {
S[u] = max (max(S[lson(u)], S[rson(u)]), R[lson(u)] + L[rson(u)]);
L[u] = L[lson(u)] + (L[lson(u)] == length(lson(u)) ? L[rson(u)] : 0);
R[u] = R[rson(u)] + (R[rson(u)] == length(rson(u)) ?

R[lson(u)] : 0);
}

void build (int u, int l, int r) {
lc[u] = l;
rc[u] = r;

if (l == r) {
maintain(u, 1);
return;
}

int mid = (l + r) / 2;
build(lson(u), l, mid);
build(rson(u), mid + 1, r);
pushup(u);
}

void modify (int u, int x, int v) {
if (lc[u] == x && rc[u] == x) {
maintain(u, v);
return;
}

int mid = (lc[u] + rc[u]) / 2;
if (x <= mid)
modify(lson(u), x, v);
else
modify(rson(u), x, v);
pushup(u);
}

int query (int u, int x) {
if (S[u] == length(u))
return S[u];

if (lc[u] == rc[u])
return 0;

int mid = (lc[u] + rc[u]) / 2;
if (x <= mid) {
if (x >= mid - R[lson(u)] + 1)
return R[lson(u)] + L[rson(u)];
else
return query(lson(u), x);
} else {
if (x <= mid + L[rson(u)])
return R[lson(u)] + L[rson(u)];
else
return query(rson(u), x);
}
}

int N, M;
vector<int> des;

int main () {
while (scanf("%d%d", &N, &M) == 2) {
des.clear();
build(1, 1, N);

int x;
char op[5];
while (M--) {
scanf("%s", op);
if (op[0] == 'R') {
x = des[des.size()-1];
modify(1, x, 1);
des.pop_back();
} else {
scanf("%d", &x);
if (op[0] == 'D') {
modify(1, x, 0);
des.push_back(x);
} else
printf("%d\n", query(1, x));
}
}
}
return 0;
}