题干：（Uva题不给题干了）
t组样例，每组首先给出一个M，然后给出一些线段（0 0结束），然后问怎么取能使得最少的线段覆盖区间[0, M]。
Sample Input 2 1 -1 0 -5 -3 2 5 0 0 1 -1 0 0 1 0 0 Sample Output 0 1 0 1
解题报告：
就是个贪心啊
AC代码：
#include<cstdio>#include<iostream>#include<algorithm>#include<queue>#include<map>#include<vector>#include<set>#include<string>#include<cmath>#include<cstring>#define ll long long#define pb push_back#define pm make_pair#define fi first#define se secondusing namespace std;const int MAX = 2e5 + 5;struct Node {int st,ed;Node(){}Node(int st,int ed):st(st),ed(ed){}bool operator<(const Node & b) const{if(st != b.st) return st < b.st;return ed > b.ed;}} node[MAX];int tot,n,m,cnt,nb;int ans[MAX];int main(){int t;int a,b,maxx,minn;cin>>t;while(t--) {tot=cnt=0;maxx = 0,minn = 0x3f3f3f3f;scanf("%d",&m);while(scanf("%d%d",&a,&b)) {if(a==0 && b==0) break;if(b<=0) continue;if(a>=m) continue;node[++tot] = Node(a,b);maxx = max(maxx,b);minn = min(minn,a);}sort(node+1,node+tot+1);int cure,curs;curs=cure=0;//if(minn >= m || maxx <= 0) {//puts("0");//if(t) puts("");//continue;//}int flag = 0;for(int i = 1; i<=tot; ) {if(node[i].st > curs) {break;}while(i<=tot && node[i].st<=curs) {if(node[i].ed > cure) {cure = node[i].ed;nb = i;}i++;}//if(i!=tot) i--;ans[++cnt] = nb;curs = cure;if(curs >= m) {flag=1;break;}}if(flag == 0) puts("0");else {printf("%d\n",cnt);for(int i = 1; i<=cnt; i++) printf("%d %d\n",node[ans[i]].st,node[ans[i]].ed);}if(t) puts("");}return 0 ;}总结： 那两句读数据的时候的if去掉好像也可以AC，，反正注释掉的那一部分加上也可以AC。。
