Intervals POJ - 1201 差分约束

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

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
	ll x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

struct node {
	int u, v, w;
	int nxt;
}edge[maxn];
int head[maxn];
int tot;
int dis[maxn];
bool vis[maxn];
void addedge(int u, int v, int w) {
	edge[++tot].u = u; edge[tot].v = v; edge[tot].w = w;
	edge[tot].nxt = head[u]; head[u] = tot;
}
void spfa() {
	queue<int>q;
	q.push(0);
	vis[0] = 1; dis[0] = 0;
	while (!q.empty()) {
		int u = q.front(); q.pop(); vis[u] = 0;
		for (int i = head[u]; i; i = edge[i].nxt) {
			int v = edge[i].v;
			if (dis[v] < dis[u] + edge[i].w) {
				dis[v] = dis[u] + edge[i].w;
				if (!vis[v]) {
					q.push(v); vis[v] = 1;
				}
			}
		}
	}
}

int main() {
	//ios::sync_with_stdio(0);
//	int T; rdint(T);
//	while (T--) {
		ms(head); memset(dis, -0x3f, sizeof(dis));
		ms(vis); tot = 0;
		int maxx = 0; int n;
		cin >> n;
		for (int i = 1; i <= n; i++) {
			int a, b, c; rdint(a); rdint(b); rdint(c);
			addedge(a, b + 1, c); maxx = max(maxx, b + 1);
		}
		for (int i = 1; i <= maxx; i++) {
			addedge(i, i - 1, -1); addedge(i - 1, i, 0);
		}
		spfa();
		cout << dis[maxx] << endl;
//		if (T)cout << endl;
//	}
	return 0;
}