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【POJ 1201】Intervals（差分约束+SPFA）

2016-11-10 21:39 603 查看
Intervals

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 25721 Accepted: 9843
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
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【题解】【差分约束】
【约束条件为 a[y+1]-a[x]>=c[i],a[x+1]-a[x]<=1,a[x+1]-a[x]>=0,全部转化为>=，跑最长路即可】


#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[500010],next[500010],p[50010],val[500010],tot;
int dis[500010],minn=0x7fffffff,maxn,ans,n;
bool b[500010];

inline void add(int x,int y,int z)
{
tot++; a[tot]=y; next[tot]=p[x]; p[x]=tot; val[tot]=z;
return ;
}
inline int spfa(int s,int t)
{
memset(dis,-1,sizeof(dis));
memset(b,true,sizeof(b));
queue<int>que;
que.push(s); b[s]=false; dis[s]=0;
while (!que.empty())
{
int u,v;
u=que.front(); que.pop();
v=p[u]; b[u]=true;
while (v!=0)
{
if (dis[a[v]]<dis[u]+val[v])
{
dis[a[v]]=dis[u]+val[v];
if (b[a[v]])
{
b[a[v]]=false;
que.push(a[v]);
}
}
v=next[v];
}
}
return dis[t];
}
int main()
{
int i,j;
scanf("%d",&n);
for (i=1;i<=n;++i)
{
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
minn=min(x,minn); maxn=max(y+1,maxn);
add(x,y+1,z);
}
for (i=minn;i<maxn;++i)
{add(i+1,i,-1); add(i,i+1,0);}
ans=spfa(minn,maxn);
printf("%d",ans);
return 0;
}
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标签：  图 最短路 数学相关 差分约束
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