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【区间覆盖问题】uva 10020 - Minimal coverage
可以说是区间覆盖问题的例题...
Note:
区间包含+排序扫描；
要求覆盖区间[s, t];
1、把各区间按照Left从小到大排序，如果区间1的起点大于s，则无解（因为其他区间的左起点更大）；否则选择起点在s的最长区间;
2、选择区间[li, ri]后，新的起点应更新为ri，并且忽略所有区间在ri之前的部分；
| Minimal coverage |
The Problem
Given several segments of line (int the X axis) with coordinates [Li,Ri]. You are to choose the minimal amount of them, such they would completely cover the segment [0,M].
The Input
The first line is the number of test cases, followed by a blank line.
Each test case in the input should contains an integer M(1<=M<=5000), followed by pairs "Li Ri"(|Li|, |Ri|<=50000, i<=100000), each on a separate line. Each test case of input is terminated by pair "0 0".
Each test case will be separated by a single line.
The Output
For each test case, in the first line of output your programm should print the minimal number of line segments which can cover segment [0,M]. In the following lines, the coordinates of segments, sorted by their left end (Li), should be printed in the same format as in the input. Pair "0 0" should not be printed. If [0,M] can not be covered by given line segments, your programm should print "0"(without quotes).
Print a blank line between the outputs for two consecutive test cases.
Sample Input
21-1 0-5 -32 50 01-1 00 10 0
Sample Output
010 1
代码如下
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 #include<algorithm> 6 using namespace std; 7 const int maxn = 100010; 8 struct Seg 9 {10 int L, R;11 bool operator < (const Seg &a) const12 {13 if(L != a.L) return L < a.L;14 else return R > a.R;15 }16 } seg[maxn], ans[maxn];17 void solve(int n, int m)18 {19 int cnt = 0, l = 0, r = 0;//起点l，起点l最长区间的右端点r20 if(seg[0].L > l)21 {22 printf("0\n");23 return ;24 }//无解25 bool ok = false;26 while(1)27 {28 if(l >= m) break;29 int i, pos = 0;30 ok = false;//判断变量ok，重要！31 for(i = 0; i < n; i++)32 {33 if(seg[i].L <= l)34 {35 if(seg[i].R > r)36 {37 pos = i;38 r = seg[pos].R;39 ok = true;40 }//寻找起点在l的最长区间41 }42 }43 if(ok)44 {45 l = r;46 ans[cnt].L = seg[pos].L;47 ans[cnt++].R = seg[pos].R;48 }//更新起点l49 else break;50 }51 if(ok)52 {53 printf("%d\n", cnt);54 for(int i = 0; i < cnt; i++)55 printf("%d %d\n", ans[i].L, ans[i].R);56 }57 else printf("0\n");58 }59 int main()60 {61 int T;62 scanf("%d", &T);63 for(int kase = 0; kase < T; kase++)64 {65 if(kase) printf("\n");66 memset(seg, 0, sizeof(seg));67 memset(ans, 0, sizeof(ans));68 int m; scanf("%d", &m);69 int i = 0, L, R;70 while(scanf("%d%d", &L, &R))71 {72 if(!L && !R) break;73 seg[i].L = L; seg[i].R = R;74 i++;75 }76 sort(seg, seg+i);//排序77 solve(i, m);78 }79 return 0;80 }
【区间覆盖问题】uva 10020 - Minimal coverage