UVa 10020 - Minimal coverage

链接：

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=113&page=show_problem&problem=961

类型： 贪心

原题：

The Problem

Given several segments of line (int the X axis) with coordinates [L_i,R_i]. 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 "L_i R_i"(|L_i|, |R_i|<=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 (L_i), 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

2

1
-1 0
-5 -3
2 5
0 0

1
-1 0
0 1
0 0

Sample Output

0

1
0 1

分析与总结：

贪心的经典区间最小覆盖问题。LRJ《算法入门经典》P154。 

对于区间【left, right】, 按照left从小到大排序，设起点为s， 那么选择left<=s中的right最大的那个为新的起点s， 如此循环下去，一直到起点s大于目标长度为止。

/* 
 *   UVa: 10020 - Minimal coverage
 *   Result: Accept
 *   Time: 0.080s
 *   Author: D_Double                                                                 
 */
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#define MAXN 100005
using namespace std;
int M;

struct Segment{
	int left;
	int right;
	friend bool operator < (const Segment&a, const Segment&b){
		if(a.left!=b.left) return a.left<b.left;
		return a.right>b.left;
	}
}arr[MAXN];
int nIndex;

int ans[MAXN], N;

int main(){
	freopen("input.txt","r",stdin);
	int T, i;
	scanf("%d",&T);
	while(T--){
		nIndex=0;
		scanf("%d",&M);
		while(~scanf("%d %d",&arr[nIndex].left, &arr[nIndex].right)){
			if(!arr[nIndex].left && !arr[nIndex].right) break;
			if(arr[nIndex].right>0) ++nIndex;
		}

		sort(arr, arr+nIndex);
		int cur = 0;
		N=-1;

		bool flag=false;

		if(arr[0].left > 0){
			printf("0\n");
			if(T) printf("\n");
			continue;
		}

		for(i=0; i<nIndex; ++i){
			if(arr[i].left <= cur){
				if(N==-1) 
					ans[++N]=i;
				else if(arr[i].right > arr[ans[N]].right)
					ans[N]=i;
			}
			else{
				cur=arr[ans[N]].right;
				ans[++N]=i;
			}
			if(arr[ans[N]].right>=M){
				flag=true; 
				break;
			}
		}
 
		if(flag){
			printf("%d\n", N+1);
			for(i=0; i<=N; ++i)
				printf("%d %d\n", arr[ans[i]].left, arr[ans[i]].right);
		}
		else
			printf("0\n");
		if(T) printf("\n");
	}
	return 0;
}

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        原创 http://blog.csdn.net/shuangde800 ， By   D_Double  (转载请标明)