zoj 1610 Count the Colors

类型：一维线段树

来源：ZOJ Monthly, May 2003

题目：线段染色问题，求最后最上面的颜色段数

思路：对每一次染色执行update操作，当颜色未完全覆盖当前段时，当前根的颜色块需要下移！然后执行一次query操作记录所有线段的颜色

数据：

2

1 2 3

4 5 3

结果为2

// zoj 1610// tle wa wa wa ac 90ms#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define FOR(i,a,b) for(i = (a); i < (b); ++i)#define FORE(i,a,b) for(i = (a); i <= (b); ++i)#define CLR(a,b) memset(a,b,sizeof(a))const int MAXN = 8010;const int MAXM = 8010;int n, a, b, c;int tree[4 * MAXN], cnt[MAXM], col[MAXM];void update(int root, int l, int r, int x1, int x2) {    if(l == x1 && r == x2) {        tree[root] = c;        return ;    }    if(l == r - 1)        return ;    if(tree[root] != -1) {        tree[root << 1] = tree[root];        tree[root << 1 | 1] = tree[root];        tree[root] = -1;    }    int mid = (l + r) >> 1;    if(mid >= x2)        update(root << 1, l, mid, x1, x2);    else if(mid <= x1)        update(root << 1 | 1, mid, r, x1, x2);    else {        update(root << 1, l, mid, x1, mid);        update(root << 1 | 1, mid, r, mid, x2);    }}void query(int root, int l, int r) {    int i;    if(tree[root] != -1) {        FOR(i, l, r)            col[i] = tree[root];        return ;    }    if(l == r - 1)        return ;    int mid = (l + r) >> 1;    query(root << 1, l, mid);    query(root << 1 | 1, mid, r);}int main() {    int i;    while(scanf("%d", &n) != EOF) {        CLR(tree, -1), CLR(cnt, 0), CLR(col, -1);        int maxc = -1, maxb = -1;        FOR(i, 0, n) {            scanf("%d %d %d", &a, &b, &c);            maxc = max(maxc, c);            maxb = max(maxb, b);            update(1, 0, 8001, a, b);        }        int tmp = -1;        query(1, 0, 8001);        FOR(i, 0, maxb) {            if(col[i] == -1) {                tmp = -1;                continue;            }            if(col[i] != tmp)                ++cnt[col[i]], tmp = col[i];        }        FORE(i, 0, maxc)            if(cnt[i] != 0)                printf("%d %d\n", i, cnt[i]);        printf("\n");    }    return 0;}/*21 4000 14000 6000 021 2 34 5 3*/