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;
}

ural 1019 Line Painting(线段树)