poj 1201 Intervals & 1716 Integer Intervals 差分约束

       题目是求一个序列，要满足给定条件，求它的最少个数

       设d[i]为在[0, i)区间内答案的个数，其中d[0] = 0

       则对题目输入的a, b ,c有：d[b+1] – d[a] >= c

       题意隐藏的不等式有： 0 <= d[i+1] – d[i] <= 1, d[i] – d[0] >= 0

       因为是求最少个数，所以 以0为源点求最长路，答案就为d[n] – d[1]

 

      1716，则把上面的c改为2即可

 

#include <iostream>
#include <vector>
#include <queue>
#include <cstring>
#include <cstdio>
using namespace std;

const int MAX = 50005;
const int INF = 1000000000;

struct edge
{
	int v;
	int cost;
	edge(){}
	edge(int v, int cost)
	{
		this->v = v;
		this->cost = cost;
	}
};

vector<edge> adj[MAX];
queue<int> Q;
bool is_in_queue[MAX];
int dis[MAX];
int _min, _max;
int n;
int cnt[MAX];

void spfa()
{
	while (!Q.empty()) Q.pop();
	int u;
	Q.push(_min);

	memset(is_in_queue, false, sizeof(is_in_queue));
	for (int i = _min; i <= _max+1; i++)
		dis[i] = -INF;

	dis[_min] = 0;
	is_in_queue[_min] = true;
	while (!Q.empty())
	{
		u = Q.front();
		Q.pop();
		is_in_queue[u] = false;

		for (int i = 0; i < adj[u].size(); i++)
		{
			int v = adj[u][i].v;
			int w = adj[u][i].cost;
			if (dis[v] < dis[u] + w)
			{
				dis[v] = dis[u]+w;

				if (!is_in_queue[v])
				{
					Q.push(v);
					is_in_queue[v] = true;
				}		
			}
		}
	}
}
int main()
{
	int a, b, c;
	while (cin >> n)
	{
		_min = INF;
		_max = -1;

		for (int i = 0; i < n; i++)
		{
			scanf("%d%d%d", &a, &b, &c);
			adj[a].push_back(edge(b+1, c));
			_min = min(_min, a);
			_max = max(_max, b);
		}

		for (int i = _min; i <= _max; i++)
		{
			adj[0].push_back(edge(i, 0));
			adj[i].push_back(edge(i+1, 0));
			adj[i+1].push_back(edge(i, -1));
		}

		adj[0].push_back(edge(_max+1, 0));

		spfa();

		cout << dis[_max+1] - dis[_min] << endl;

		for (int i = 0; i <= _max+1; i++)
			adj[i].clear();
	}
	return 0;
}