Intervals Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 18456 Accepted: 6919 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. 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 Source Southwestern Europe 2002 考差分約束，自我感覺是這類問題是找不等式關系，然後想方法建圖，然後用一遍spfa就可以了 初步接觸差分約束 [cpp] #include <stdio.h> #include <string.h> #include <math.h> struct num { int end,val,next; }a[1000000]; int b[50100]; int d[50100],queue[1000000]; int status[50100]; int INF=0x7fffff; int main() { int spfa(int sta,int end); int i,j,n,m,s,t; int x,y,val,max,min; while(scanf("%d",&n)!=EOF) { memset(b,-1,sizeof(b)); max=-1; min=INF; for(i=0,j=0;i<=n-1;i++) { scanf("%d %d %d",&x,&y,&val); x++; y++; if(x-1<min) { min=x-1; } if(y>max) { max=y; } a[j].end= y; a[j].val=val; a[j].next=b[x-1]; b[x-1]=j; j++; } for(i=min+1;i<=max;i++) { a[j].end=i; a[j].val=0; a[j].next=b[i-1]; b[i-1]=j; j++; a[j].end=i-1; a[j].val=-1; a[j].next=b[i]; b[i]=j; j++; } t=spfa(min,max); printf("%d\n",t); } return 0; } int spfa(int sta,int end) { int i,j,top,base,x,xend,y; for(i=sta;i<=end;i++) { d[i]=-1*INF; } top=base=0; queue[top++]=sta; d[sta]=0; memset(status,0,sizeof(status)); status[sta]=1; while(base<top) { x=queue[base++]; status[x]=0; for(j=b[x];j!=-1;j=a[j].next) { y=a[j].end; if(d[y]<(d[x]+a[j].val)) { d[y]=d[x]+a[j].val; if(!status[y]) { queue[top++]=y; } } } } return (d[end]); }
