poj 1716 Integer Intervals （差分约束）

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

4

3 6

2 4

0 2

4 7

Sample Output

4

题解

这题比考试题简单。。。轻松混过！跟着大佬的步伐水水水（不大佬几乎不做水题，只有这几道我能勉强做出来。。）

代码

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

struct Edge {
int v, w, next;
} e[200010];
int n = 0, t = 0x3fffff, m, num = 0, a[100010], h[200010];

void add(int u, int v, int w) {
num ++;
e[num].v = v;
e[num].w = w;
e[num].next = h[u];
h[u] = num;
}

int head = 0, tail = 1, vis[100010], dis[100010], queue[500010];
void spfa(int s) {
memset(dis, -63, sizeof(dis));
queue[1] = s; vis[s] = 1; dis[s] = 0;
while(head < tail) {
int u = queue[++ head]; vis[u] = false;
for(int i = h[u]; i; i = e[i].next) {
int v = e[i].v;
if(dis[v] < dis[u] + e[i].w) {
dis[v] = dis[u] + e[i].w;
if(! vis[v]) {
vis[v] = true;
queue[++ tail] = v;
}
}
}
}
}
int main() {
scanf("%d", &m);
for(int i = 1; i <= m; i ++) {
int l, r;
scanf("%d %d", &l, &r);
n = max(n, r + 1); t = min(t, l); add(l, r + 1, 2);
}
for(int i = t; i < n; i ++)
add(i + 1, i, -1), add(i, i + 1, 0);
spfa(t);
printf("%d", dis
);
return 0;
}