POJ1201 Intervals <差分约束系统>

Intervals

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

标签：差分约束系统

题目大意：给出n个区间，试确定一个集合使得对于i=1..n，第i个区间内至少有ci个数，并使得此集合尽量小，输出最小大小。

题解：
首先用前缀和。sum[i]表示从1到i中共选出多少个数到集合中。这样对于集合[ai,bi]，有sum[bi]-sum[ai-1]>=ci，于是我们可以从点ai-1到bi连一条权值为ci的边。因为题意是要满足所有的边，所以我们需要找最长路。
此题有一些细节问题。首先，找最长路需要起点和终点，我们需要找到这些集合覆盖的范围，即找到左端点（其实应该是左端点-1）的最小值s和右端点的最大值t，找s到t的最大值。此外，光有上述的那些边时无法构成一个连通图的，所以我们需要找一些隐含条件。可以发现有sum[i]-sum[i-1]>=0，sum[i]-sum[i-1]<=1，为了保持一致，应将后面的式子转化为大于等于，即sum[i-1]-sum[i]>=-1，这样对于i=s+1~t，从i-1到i连接一条权值为0的路，从i到i-1连接一条权值为-1的路，之后就可以直接用SPFA找最长路了。

最后附上AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define MAX_N 50000
using namespace std;
int n, cnt, s, t, pre[MAX_N+5], dis[MAX_N+5];
struct node {int v, c, nxt;} E[MAX_N*3+5];
void insert(int u, int v, int c) {
	E[++cnt].v = v, E[cnt].c = c;
	E[cnt].nxt = pre[u], pre[u] = cnt;
}
void SPFA() {
	queue <int> que;
	bool inque[MAX_N+5];
	memset(dis, 128, sizeof(dis));
	memset(inque, 0, sizeof(inque));
	dis[s] = 0;
	que.push(s), inque[s] = true;
	while (!que.empty()) {
		int u = que.front();
		for (int i = pre[u]; i; i = E[i].nxt) {
			int v = E[i].v, c = E[i].c;
			if (dis[u]+c > dis[v]) {
				dis[v] = dis[u]+c;
				if (!inque[v])	que.push(v), inque[v] = true;
			}
		}
		que.pop(), inque[u] = false;
	}
}
int main() {
	while (scanf("%d", &n) != EOF) {
		cnt = 0, s = 50000, t = 0;
		memset(pre, 0, sizeof(pre));
		memset(E, 0, sizeof(E));
		for (int i = 0; i < n; i++) {
			int u, v, c;
			scanf("%d%d%d", &u, &v, &c);
			insert(u, v+1, c);
			s = min(s, u);
			t = max(t, v+1);
		}
		for (int i = s; i < t; i++)	insert(i, i+1, 0), insert(i+1, i, -1);
		SPFA();
		printf("%d\n", dis[t]);
	}
	return 0;
}