【POJ 1201】 Intervals(差分约束系统)
2024-08-31 03:26:43
11
1716的升级版 把原本固定的边权改为不固定。
Intervals
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 23817 | Accepted: 9023 |
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.
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
关于差分约束系统,百度各种大牛博客讲的都非常具体,简单说就是通过不等关系建立约束系统图。然后跑最短路(大于关系则跑最长路)
回到此题,题目要求找出一个最小集合S。满足对于n个范围[ai,bi],S中存在ci个及ci个以上不同的点在范围内
令Zi表示满足条件的情况下,0~i点至少有多少点在集合内
则Zbi-Zai >= ci
仅仅有这一个条件构造出来的图可能不是全然连通的,所以须要找一些“隐含条件”
不难发现 对于相邻的点 0 <= Zi-Z(i-1) <= 1 保证关系符同样 转化为
Zi-Z(i-1) >= 0
Z(i-1)-Zi >= -1
用这三个关系,就可以构造差分约束系统,然后SPFA或者Bellman跑一趟最长路(满足全部条件)
代码例如以下:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue> using namespace std;
const int INF = 0x3f3f3f3f;
const int msz = 1e5;
const int mod = 1e9+7;
const double eps = 1e-8; struct Edge
{
int v,w,next;
}; Edge eg[233333];
int head[50050];
bool vis[50050];
int dis[50050];
int tp,st,en; void Add(int u,int v,int w)
{
eg[tp].v = v;
eg[tp].w = w;
eg[tp].next = head[u];
head[u] = tp++;
} int SPFA()
{
memset(vis,0,sizeof(vis));
memset(dis,-INF,sizeof(dis));
queue <int> q;
dis[st] = 0;
vis[st] = 1;
int u,v,w; q.push(st); while(!q.empty())
{
u = q.front();
q.pop();
vis[u] = 0;
for(int i = head[u]; i != -1; i = eg[i].next)
{
v = eg[i].v;
w = eg[i].w;
if(dis[v] < dis[u]+w)
{
dis[v] = dis[u]+w;
if(!vis[v])
{
q.push(v);
vis[v] = 1;
}
}
}
}
return dis[en];
} int main(int argc,char **argv)
{
int n;
int u,v,w; while(~scanf("%d",&n))
{
tp = 0;
memset(head,-1,sizeof(head)); en = 0,st = INF; while(n--)
{
scanf("%d%d%d",&u,&v,&w);
Add(u,v+1,w);
en = max(en,v+1);
st = min(st,u);
} for(int i = st; i < en; ++i)
{
Add(i,i+1,0);
Add(i+1,i,-1);
}
printf("%d\n",SPFA());
}
return 0;
}
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