ural 1019 Line Painting(线段树)

题目链接：ural 1019 Line Painting

题目大意：一个0~1e9的区间，初始都是白的，现进行N次操作，每次将一段区间图上一中颜色。最后问说连续最长的白色区间。

解题思路：线段树区间合并,每个节点即维护一个区间，很经典。注意坐标需要离散化，但是还是要将0和1e9放进去。

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

using namespace std;

const int maxn = 10005;
const int INF = 0x3f3f3f3f;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
struct pii {
int len, pos;
pii (int len = 0, int pos = 0) {
this->len = len;
this->pos = pos;
}
friend bool operator < (const pii& a, const pii& b) {
if (a.len != b.len)
return a.len < b.len;
return a.pos > b.pos;
}
}s[maxn << 2];
int N, M, pos[maxn], cnt = 0;
int lc[maxn << 2], rc[maxn << 2], L[maxn << 2], R[maxn << 2], set[maxn << 2];

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

inline void maintain(int u, int w) {
set[u] = w;
if (w) {
L[u] = R[u] = length(u);
s[u] = pii(R[u], pos[lc[u]]);
} else {
L[u] = R[u] = 0;
s[u] = pii(0, INF);
}
}

inline void pushup(int u) {
pii cur (L[rson(u)] + R[lson(u)], pos[lc[rson(u)]] - R[lson(u)]);
s[u] = max(max(s[lson(u)], s[rson(u)]), cur);
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);
//printf("%d %d:%d %d!\n", lc[u], rc[u], s[u].pos, s[u].pos + s[u].len);
}

inline void pushdown(int u) {
if (set[u] != -1) {
maintain(lson(u), set[u]);
maintain(rson(u), set[u]);
set[u] = -1;
}
}

void build(int u, int l, int r) {
lc[u] = l; rc[u] = r;
set[u] = -1;

if (l == r) {
L[u] = R[u] = length(u);
s[u] = pii(R[u], pos[lc[u]]);
return;
}
int mid = (lc[u] + rc[u]) >> 1;
build(lson(u), l, mid);
build(rson(u), mid+1, r);
pushup(u);
}

void modify(int u, int l, int r, int w) {
if (l <= lc[u] && rc[u] <= r) {
maintain(u, w);
return;
}

pushdown(u);
int mid = (lc[u] + rc[u]) >> 1;
if (l <= mid)
modify(lson(u), l, r, w);
if (r > mid)
modify(rson(u), l, r, w);
pushup(u);
}

pii query(int u, int l, int r) {
if (l <= lc[u] && rc[u] <= r)
return s[u];
pushdown(u);
int mid = (lc[u] + rc[u]) >> 1;
pii ret(0, INF);
if (l <= mid)
ret = max(ret, query(lson(u), l, r));
if (r > mid)
ret = max(ret, query(rson(u), l, r));
pushup(u);
return ret;
}

struct Seg {
int l, r;
char op[5];
void read() { scanf("%d%d%s", &l, &r, op);}
}q[maxn];

int find (int x) {
return lower_bound(pos, pos + M, x) - pos;
}

void init () {
int n = 0;
for (int i = 1; i <= N; i++) {
q[i].read();
pos[i*2-1] = q[i].l;
pos[i*2] = q[i].r;
}

sort(pos + 1, pos + 1 + N * 2);
M = unique(pos + 1, pos + 1 + N * 2) - (pos+1);
pos[0] = 0;
pos[M + 1] = 1e9;
build(1, 0, M);
}

int main () {
while (scanf("%d", &N) == 1) {
init();

for (int i = 1; i <= N; i++) {
modify(1, find(q[i].l), find(q[i].r)-1, q[i].op[0] == 'w' ? 1 : 0);
}
printf("%d %d\n", s[1].pos, s[1].pos + s[1].len);
}
return 0;
}