POJ 1716 Integer Intervals 【差分约束】

最新推荐文章于 2018-11-05 10:03:51 发布

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Integer Intervals

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 15647		Accepted: 6631

Description

An integer interval [a,b], a < b, is a set of all consecutive integers beginning with a and ending with b.
Write a program that: finds the minimal number of elements in a set containing at least two different integers from each interval.

Input

The first line of the input contains the number of intervals n, 1 <= n <= 10000. Each of the following n lines contains two integers a, b separated by a single space, 0 <= a < b <= 10000. They are the beginning and the end of an interval.

Output

Output the minimal number of elements in a set containing at least two different integers from each interval.

Sample Input

Sample Output

4
建图不要用vector，这道题会T；

#include<cstdio>
#include<queue>
#include<vector>
#include<cstring>
using namespace std;
const int MAX = 100007;
const int INF = 0x3f3f3f3f;
struct node{
    int to, c, next;
    node() {}
    node(int to, int c, int next) : to(to), c(c), next(next) {}
}e[MAX];
int head[MAX];
int dis[MAX], vis[MAX];
queue <int> q;
int cnt = 0;
void add(int x, int y, int z){
    e[cnt] = {y, z, head[x]};
    head[x] = cnt++;
}
int mn, mx;
int spfa()
{
    fill(dis + mn + 1, dis + mx + 1, -INF);
    vis[mn] = 1;
    q.push(mn);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        vis[u] = 0;
        for(int i = head[u]; i != -1; i = e[i].next)
        {
            int to = e[i].to;
            int c = e[i].c;
            if(dis[to] < dis[u] +c)
            {
                dis[to] = dis[u] + c;
                if(!vis[to])
                {
                    q.push(to);
                    vis[to] = 1;
                }
            }
        }
    }
    return dis[mx];
}
int main(){
    int n;
    scanf("%d", &n);
    memset(head, -1, sizeof head);
    mx = -INF, mn = INF;
    for(int i = 0, x, y; i < n; i++){
        scanf("%d%d", &x, &y);
        y++;
        add(x, y, 2);
        mn = min(x, mn), mx = max(mx, y);
    }
    for(int i = mn; i <= mx; i++){
        add(i, i + 1, 0);
        add(i + 1, i, -1);
    }
    printf("%d\n", spfa());
    return 0;
}