题目链接：
http://acm.hit.edu.cn/hoj/problem/view?id=2244
题目描述：
Get the Colors
Submitted : 579, Accepted : 111
This problem arises from a research on the newest Web Data Search & Mining technology.
There are several points on the x-axis, each with a color and a unique x-coordinate.
Your task is to calculate the minumum interval on the x-axis that contains all the colors.
Input
This problem has multiple test cases. Each test case begins with an integer N that specifies the number of points. N lines follow, each with two integers Xi and Ci, specifying point i‘s x-coordinate and color. 1 ≤ N ≤ 10000, 1 ≤ Ci ≤ 1000. Xi will fit in signed 32-bit integer.
Output
For each test case, output a single integer, which is the length of the minimum interval that contains all the colors.
Sample Input
6 -5 3 -3 1 0 2 1 3 5 2 10 1
Sample Output
4
题目大意：
给x轴上点的坐标与颜色，求出包含全部颜色的点的最小区间长度
思路：
先算出共有多少种颜色（cnt种），然后预处理next数组，计算出每一个点相同颜色的下一个位置是多少
然后从头开始扫，把位置坐标插入优先队列中（保证队列中只有cnt个点，且他们颜色不同）每次pop出坐标最小的点，将next点push入队
这样每次算队列中最大坐标与最小坐标之差
最小的即为答案
代码：
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <queue>
6 using namespace std;
7
8 typedef long long LL;
9
10 const int N = 10010;
11 const int M = 1010;
12 const LL INF = 1e15;
13
14 struct Node {
15 LL x; //坐标
16 int c; //颜色
17 Node(LL x = -1, int c = 0) :x(x), c(c) {}
18 const bool operator < (const Node& A) const {
19 return x < A.x;
20 }
21 }no[N];
22
23 struct State {
24 LL p; //坐标
25 int x; //数组下标
26 State(LL p = 0, int x = 0) :p(p), x(x) {}
27 const bool operator < (const State& A) const {
28 return p > A.p;
29 }
30 };
31
32 int n, cnt, ne[N], po[M];
33 bool vis[M];
34
35 int main() {
36 while (cin >> n) {
37 cnt = 0;
38 memset(vis, 0, sizeof(vis));
39 for (int i = 0; i < n; ++i) {
40 scanf("%lld%d", &no[i].x, &no[i].c);
41 if (!vis[no[i].c])vis[no[i].c] = true, ++cnt; //记录颜色数目
42 }
43 sort(no, no + n); //按坐标排序
44 int tmp = 0, t = 0;
45 LL best = 0, ans = INF;
46 memset(vis, 0, sizeof(vis));
47 memset(po, 0, sizeof(po));
48 while (t < n) {
49 if (!vis[no[t].c])vis[no[t].c] = true, ++tmp;
50 po[no[t].c] = t;
51 if (tmp >= cnt)break;
52 ++t;
53 } //t为第一个全部颜色都包含的位置
54 priority_queue<State> que;
55 for (int i = 0; i < M; ++i)if (vis[i]) {
56 que.push(State(no[po[i]].x, po[i]));
57 best = max(best, no[po[i]].x); //队列中坐标最大值
58 }
59 bool hasnext[N] = { 0 };
60 for (int i = t + 1; i < n; ++i) { //预处理next数组，hasnext记录是否为尾
61 hasnext[po[no[i].c]] = true;
62 ne[po[no[i].c]] = i;
63 po[no[i].c] = i;
64 }
65 while (!que.empty()) {
66 State tmp = que.top();
67 que.pop();
68 ans = min(best - tmp.p, ans); //记录区间长
69 if (!hasnext[tmp.x])break;
70 tmp = State(no[ne[tmp.x]].x, ne[tmp.x]);
71 best = max(best, tmp.p); //维护队中坐标最大值
72 que.push(tmp);
73 }
74 printf("%lld\n", ans);
75 }
76 }