标签：lld ons tput struct follow static problem tar ble
题目链接：
http://acm.hit.edu.cn/hoj/problem/view?id=2244
题目描述：
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 1Sample 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 }
标签：lld ons tput struct follow static problem tar ble
原文地址：http://www.cnblogs.com/hyp1231/p/6984233.html
