【线段树区间更新】Count the Colors ZOJ

Think:
1知识点：线段树区间更新
2题意：一条长度为8000的布条，每次输入xi, yi, ci, 表示将区间[xi, yi]染色为ci，询问最终状态每个颜色分别染了几段，按照index顺序输入
3反思：
1>左右儿子结点编号不要传入错误
4思考：先将区间[1, 8000]全部初始化为-1, 之后输入xi, yi, ci之后将区间[xi+1, yi]颜色更新为ci, 更新完成后，进行区间[1, 8000]的颜色信息查询，将符合要求的每一段颜色信息存储到num数组中，最后进行总的判断输入即可

vjudge题目链接

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 8014;struct Tree{    int c;}tree[N<<2];int tmp, num[N];void add_v(int L, int R, int c, int l, int r, int rt);/*区间更新操作:将区间[L, R]颜色置为c*/void query(int l, int r, int rt);/*查询区间[l, r]的颜色信息，将需要信息存储到num数组中*/void down(int rt);/*将当前结点颜色信息down到左右儿子结点颜色信息*/int main(){    int n, i, x, y, c;    while(~scanf("%d", &n)){        memset(tree, -1, sizeof(tree));        memset(num, 0, sizeof(num));        for(i = 0; i < n; i++){            scanf("%d %d %d", &x, &y, &c);            if(x < y)                add_v(x+1, y, c, 1, 8000, 1);/*区间查询从1开始，存储区间左结点右移一个单位*/        }        tmp = -1;        query(1, 8000, 1);        for(i = 0; i <= 8000; i++){            if(num[i])                printf("%d %d\n", i, num[i]);        }        printf("\n");    }    return 0;}void add_v(int L, int R, int c, int l, int r, int rt){    if(L <= l && r <= R){        tree[rt].c = c;        return;    }    down(rt);    int mid = (l+r)/2;    if(L <= mid)        add_v(L, R, c, l, mid, rt<<1);    if(R > mid)        add_v(L, R, c, mid+1, r, rt<<1|1);}void down(int rt){    if(tree[rt].c != -1){        tree[rt<<1].c = tree[rt<<1|1].c = tree[rt].c;        tree[rt].c = -1;    }}void query(int l, int r, int rt){    if(l == r){        if(tree[rt].c != -1 && tree[rt].c != tmp)            num[tree[rt].c]++;        tmp = tree[rt].c;        return;    }    down(rt);    int mid = (l+r)/2;    query(l, mid, rt<<1);    query(mid+1, r, rt<<1|1);}