Problem Description
You are given n closed, integer intervals [ai, bi] and n integers c1, ..., cn.
Write a program that:
reads the number of intervals, their end points and integers c1, ..., cn from the standard input,
computes the minimal size of a set Z of integers which has at least ci common elements with interval [ai, bi], for each i=1,2,...,n,
writes the answer to the standard output.
Write a program that:
reads the number of intervals, their end points and integers c1, ..., cn from the standard input,
computes the minimal size of a set Z of integers which has at least ci common elements with interval [ai, bi], for each i=1,2,...,n,
writes the answer to the standard output.
Input
The first line of the input contains an integer n (1 <= n <= 50000) -- the number of intervals. The following n lines describe the intervals. The (i+1)-th line of the input contains three integers ai, bi and ci separated by single spaces and such that 0 <= ai <= bi <= 50000 and 1 <= ci <= bi - ai+1.
Output
The output contains exactly one integer equal to the minimal size of set Z sharing at least ci elements with interval [ai, bi], for each i=1,2,...,n.
Sample Input
5
3 7 3
8 10 3
6 8 1
1 3 1
10 11 1
Sample Output
6
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差分约束n个数和m个约束条件，求满足条件的一组数据，其中xi-xj<=bs正好满足spfa中的三角式子，于是该题就变成了图论题即求最短路的问题，找范围（找到一个源点）建图时要建全。
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1 #include<iostream> 2 #include<string> 3 #include<cstring> 4 #include<cstdio> 5 #include<cctype> 6 #include<queue> 7 #include<stack> 8 using namespace std; 9 const int inf=999999999; 10 const int maxn=50050; 11 int dis[maxn],vis[maxn],n; 12 int aa,bb; 13 struct node 14 { 15 int v; 16 int value; 17 }tmp; 18 vector<node>map[maxn]; 19 void spfa()//求最长路 20 { 21 memset(vis,0,sizeof(vis)); 22 queue<int>Q; 23 for(int it=aa;it<=bb;it++) 24 dis[it]=-inf; 25 while(!Q.empty()) 26 Q.pop(); 27 Q.push(aa); 28 vis[aa]=1; 29 dis[aa]=0; 30 while(!Q.empty()) 31 { 32 int gs=Q.front(); 33 Q.pop(); 34 vis[gs]=0; 35 int sn=map[gs].size(); 36 for(int it=0;it<sn;it++) 37 { 38 if((dis[map[gs][it].v]<dis[gs]+map[gs][it].value)) 39 { 40 dis[map[gs][it].v]=dis[gs]+map[gs][it].value; 41 if(vis[map[gs][it].v]==0) 42 { 43 Q.push(map[gs][it].v); 44 vis[map[gs][it].v]=1; 45 } 46 } 47 } 48 } 49 } 50 int main() 51 { 52 scanf("%d",&n); 53 int st,en,w; 54 aa=maxn; 55 bb=-1; 56 //满足不等式s[en]-s[st-1]>=w; 57 for(int i=1;i<=n;i++) 58 { 59 scanf("%d %d %d",&st,&en,&w); 60 if(st<aa)aa=st; 61 if((en+1)>bb)bb=en+1; 62 tmp.value=w; 63 tmp.v=en+1; 64 map[st].push_back(tmp); 65 66 } 67 //0<=s[i+1]-s[i]<=1 68 for(int it=aa;it<=bb;it++) 69 { 70 //添加0边和-1边 71 tmp.value=0; 72 tmp.v=it+1; 73 map[it].push_back(tmp); 74 tmp.value=-1; 75 tmp.v=it; 76 map[it+1].push_back(tmp); 77 } 78 spfa(); 79 printf("%d ",dis[bb]);//最大范围的最长边即为结果 80 return 0; 81 }
坚持！多刷题！