UVA Intervals

UVA Intervals

Description

• Has n sections, in the interval [ai, bi] at least mutually different ci take any integer. In seeking to meet the case of n sections, at least to take a positive integer number.

Input

• Multiple sets of data.

  A first integer and T represents the number of data lines, followed by a blank line.

  For each test:

  The first row contains an integer n (1 <= n <= 50000) representing the interval number.

  The following sections describe the row n.

  The first input (i + 1) line contains three integers ai, bi, ci, separated by spaces. Where 0 <= ai <= bi <= 50000,1 <= ci <= bi-ai + 1.

Output

• For each test, the output takes at least a number of the total number ci different integers for n interval [ai, bi].

  In addition to the last set of blank line data output.

Sample Input

1

5
3 7 3
8 10 3
6 8 1
1 3 1
10 11 1

Sample Output

6

answer:

• Differential constraints.
• When you finish Luo Gu P1250 trees do this problem, you will be shouting: "ah original title!"
• In fact, the essence is the same.
• So the solution to a problem Benpian turn essay
• Also, why is such a simple question ... purple ...

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define N 200005
using namespace std;

struct E {int next, to, dis;} e[N];
int T, m, num, n;
int h[N], dis[N];
bool vis[N];

int read()
{
    int x = 0; char c = getchar();
    while(c < '0' || c > '9') c = getchar();
    while(c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
    return x;
}

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

void spfa()
{
    queue<int> que;
    memset(dis, -0x3f, sizeof(dis));
    memset(vis, 0, sizeof(vis));
    dis[n + 1] = 0, vis[n + 1] = 1, que.push(n + 1);
    while(!que.empty())
    {
        int now = que.front();
        que.pop(); vis[now] = 0;
        for(int i = h[now]; i != 0; i = e[i].next)
            if(dis[now] + e[i].dis > dis[e[i].to])
            {
                dis[e[i].to] = dis[now] + e[i].dis;
                if(!vis[e[i].to])
                    vis[e[i].to] = 1, que.push(e[i].to);
            }
    }
}

int main()
{
    cin >> T;
    for(int dfn = 1; dfn <= T; dfn++)
    {
        n = num = 0;
        memset(h, 0, sizeof(h));
        m = read();
        for(int i = 1; i <= m; i++)
        {
            int a = read(), b = read(), c = read();
            add(a - 1, b, c), n = max(n, b);
        }
        for(int i = 1; i <= n; i++) add(i - 1, i, 0), add(i, i - 1, -1);
        for(int i = 0; i <= n; i++) add(n + 1, i, 0);
        spfa();
        printf("%d\n", dis[n]);
        if(dfn != T) printf("\n");
    }
    return 0;
}