题意：在[0, 8000]上染色n次，每次染色的区间为[x1, x2]，颜色为c，问最后每种颜色有多少段（所有数字在[0, 8000]内）。
题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=610
——>>对于端点x1, x2，对应成区间段[x1+1, x2]，接着就染染色，线段树处理一下就好了。。。#^_^
SF了几次，原因：n指染色次数，并非指在区间[0, n]上染色。
#include <cstdio>
#include <cstring>using namespace std;#define lc (o<<1)
#define rc ((o<<1)+1)const int N = 8000;
const int maxn = (8000 + 10) << 2;int col[maxn], a[maxn], cnt[maxn];void build(int o, int l, int r) {col[o] = -1;if(l == r) return;int m = (l + r) >> 1;build(lc, l, m);build(rc, m+1, r);
}void pushdown(int o) {col[lc] = col[rc] = col[o];col[o] = -1;
}void update(int o, int l, int r, int ql, int qr, int v) {if(ql <= l && r <= qr) {col[o] = v;return;}if(col[o] == v) return;if(col[o] != -1) pushdown(o);int m = (l + r) >> 1;if(ql <= m) update(lc, l, m, ql, qr, v);if(qr > m) update(rc, m+1, r, ql, qr, v);
}void query(int o, int l, int r) {if(col[o] != -1) {for(int i = l; i <= r; i++) {a[i] = col[o];}return;}if(l == r) return;int m = (l + r) >> 1;query(lc, l, m);query(rc, m+1, r);
}void solve() {memset(cnt, 0, sizeof(cnt));for(int i = 1; i <= N; i++) //遍历段if(a[i] != a[i-1])cnt[a[i]]++;for(int i = 0; i <= N; i++) //遍历颜色if(cnt[i])printf("%d %d\n", i, cnt[i]);
}int main()
{int n, x1, x2, c;while(scanf("%d", &n) == 1) {build(1, 1, N);for(int i = 0; i < n; i++) {scanf("%d%d%d", &x1, &x2, &c);update(1, 1, N, x1+1, x2, c);}memset(a, -1, sizeof(a));query(1, 1, N);solve();puts("");}return 0;
}