hdu 1540 Tunnel Warfare(线段树)

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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，每次只要判断是否在mid−R[lson(u)],mid+L[rson(u)]之间即可，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;}

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