Wooden Sticks
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9541 Accepted Submission(s): 3917
Problem Description
There is a pile of n wooden sticks. The length and weight of each stick are known in advance. The sticks are to be processed by a woodworking machine in one by one fashion. It needs some time, called setup time, for the machine to prepare processing a stick. The setup times are associated with cleaning operations and changing tools and shapes in the machine. The setup times of the woodworking machine are given as follows:
(a) The setup time for the first wooden stick is 1 minute. (b) Right after processing a stick of length l and weight w , the machine will need no setup time for a stick of length l' and weight w' if l<=l' and w<=w'. Otherwise, it will need 1 minute for setup.
You are to find the minimum setup time to process a given pile of n wooden sticks. For example, if you have five sticks whose pairs of length and weight are (4,9), (5,2), (2,1), (3,5), and (1,4), then the minimum setup time should be 2 minutes since there is a sequence of pairs (1,4), (3,5), (4,9), (2,1), (5,2).
(a) The setup time for the first wooden stick is 1 minute. (b) Right after processing a stick of length l and weight w , the machine will need no setup time for a stick of length l' and weight w' if l<=l' and w<=w'. Otherwise, it will need 1 minute for setup.
You are to find the minimum setup time to process a given pile of n wooden sticks. For example, if you have five sticks whose pairs of length and weight are (4,9), (5,2), (2,1), (3,5), and (1,4), then the minimum setup time should be 2 minutes since there is a sequence of pairs (1,4), (3,5), (4,9), (2,1), (5,2).
Input
The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case consists of two lines: The first line has an integer n , 1<=n<=5000, that represents the number of wooden sticks in the test case, and the second line contains n 2 positive integers l1, w1, l2, w2, ..., ln, wn, each of magnitude at most 10000 , where li and wi are the length and weight of the i th wooden stick, respectively. The 2n integers are delimited by one or more spaces.
Output
The output should contain the minimum setup time in minutes, one per line.
Sample Input
3
5
4 9 5 2 2 1 3 5 1 4
3
2 2 1 1 2 2
3
1 3 2 2 3 1
Sample Output
2
1
3
好吧，这是一道简单题，，，仔细点就好；
思路是；把木棍按照长度优先从小到大排好，当长度一定是按宽度从小到大排好，然后比较宽度就ok了
，具体见代码，，，，
简单易懂型：
#include<cstdio> #include<iostream> #include<algorithm> using namespace std; struct node { int l; int w; int k; }a[5005]; int main() { int T,n,j,i,temp,sum,p; scanf("%d",&T); while(T--) { sum=0; scanf("%d",&n); for(i=1;i<=n;i++) { scanf("%d%d",&a[i].l,&a[i].w); a[i].k=1;//标记 } for(i=1;i<=n;i++)//选择 why!!! { p=i; for(j=i+1;j<=n;j++) if(a[p].l>a[j].l||a[p].l==a[j].l&&a[p].w>a[j].w) //长度优先排，然后是宽度，都排好，从小到大 p=j; if(p!=i) { int t; t=a[i].l; a[i].l=a[p].l; a[p].l=t; t=a[i].w; a[i].w=a[p].w; a[p].w=t; } } // for(i=1;i<=n;i++) // printf("%d %d %d ",a[i].l,a[i].w,a[i].k); for(i=1;i<=n;i++) { if(a[i].k==1) { temp=a[i].w; for(j=i+1;j<=n;j++) { if(a[j].w>=temp&&a[j].k==1) { a[j].k=0; temp=a[j].w; sum++; } } } } printf("%d ",n-sum); } return 0; } /* 4 5 4 9 5 2 2 1 3 5 1 4 3 2 2 1 1 2 2 3 1 3 2 2 3 1 7 1 12 1 5 2 6 2 8 2 4 4 3 3 7 */
然后是更加简洁的排序sort，快排。。。
#include<cstdio> #include<iostream> #include<algorithm> using namespace std; struct node { int l; int w; int k; }a[5005]; int cmp(node a,node b) { if(a.w==b.w) return a.l<b.l; return a.w<b.w; } int main() { int T,n,j,i; scanf("%d",&T); while(T--) { scanf("%d",&n); for(i=0;i<n;i++) { scanf("%d%d",&a[i].l,&a[i].w); a[i].k=1;//标记 } sort(a,a+n,cmp); for(i=0;i<n;i++) printf("%d %d %d ",a[i].l,a[i].w,a[i].k); int num=n; int count=0; while(num) { int tl=0;int tw=0; for(i=0;i<n;i++) { if(a[i].k==1&&a[i].w>=tw&&a[i].l>=tl) { tl=a[i].l; tw=a[i].w; a[i].k=0;num--; } } count++; } printf("%d ",count); } return 0; }
稍作修改：
#include<cstdio> #include<iostream> #include<algorithm> using namespace std; struct node { int l; int w; int k; }a[5005]; int cmp(node a,node b)//从小到大排，，，对于结构体 { if(a.l==b.l) return a.w<b.w; return a.l<b.l; } int main() { int T,n,j,i,temp,sum; scanf("%d",&T); while(T--) { sum=0; scanf("%d",&n); for(i=1;i<=n;i++) { scanf("%d%d",&a[i].l,&a[i].w); a[i].k=1;//标记 } sort(a+1,a+n+1,cmp); for(i=1;i<=n;i++) printf("%d %d %d ",a[i].l,a[i].w,a[i].k); for(i=1;i<=n;i++) { if(a[i].k==1) { temp=a[i].w; for(j=i+1; j<=n; j++) { if(a[j].w>=temp&&a[j].k==1) { a[j].k=0; temp=a[j].w; sum++; } } } } printf("%d ",n-sum); } return 0; }
是的，冒泡嘛，也是可以的，，，
#include<cstdio> #include<iostream> #include<algorithm> using namespace std; struct node { int l; int w; int k; }a[5005]; int cmp(node a,node b) { if(a.l==b.l) return a.w<b.w; return a.l<b.l; } int main() { int T,n,j,i,temp,sum; scanf("%d",&T); while(T--) { sum=0; scanf("%d",&n); for(i=1;i<=n;i++) { scanf("%d%d",&a[i].l,&a[i].w); a[i].k=1;//标记 } for(i=1;i<=n;i++)//冒泡 { for(j=0;j<n-i;j++)//j从1开始就错了。。。 if(a[j].l>a[j+1].l) { int vv; vv=a[j].w; a[j].w=a[j+1].w; a[j+1].w=vv; vv=a[j].l; a[j].l=a[j+1].l; a[j+1].l=vv; } } for(i=1;i<=n;i++) printf("%d %d %d ",a[i].l,a[i].w,a[i].k); for(i=1;i<=n;i++) { if(a[i].k==1) { temp=a[i].w; for(j=i+1; j<=n; j++) { if(a[j].w>=temp&&a[j].k==1) { a[j].k=0; temp=a[j].w; sum++; } } } } printf("%d ",n-sum); } return 0; }
推荐第二种，，，，本题还可以用dp做，具体代码，有待有序更新。。。o(∩_∩)o 哈哈