POJ1201 差分约束

给定ai,bi, ci 表示区间[ai,bi]内至少有ci个点， 要求对于所有给定的ai,bi,ci，  至少多少个点才能满足题目的条件

重做这一题学到的一点是， 可以设变量来表示一些东西，然后才能找出约束的条件，  s[i]表示区间0到i内有多少个点，  那么s[bi] - s[ai-1] >= ci 就是约束的条件

当然了，也有隐藏的条件  1>= s[i] - s[i-1] >=0

可以用最长路来求这一题，最短路当然也是可以的。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <math.h>
#pragma warning(disable:4996)
#pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
using namespace std;
const int INF = <<;
const int N =  + ;
struct Node
{
    int to, dist, next;
    bool operator<(const Node&rhs)const
    {
        return dist < rhs.dist;
    }
}g[N];
int head[N], e;
int dist[N];
void addEdge(int a, int b, int c)
{
    g[e].to = b;
    g[e].dist = c;
    g[e].next = head[a];
    head[a] = e++;
}
int dij(int x, int y)
{
    priority_queue<Node> q;
    for (int i = x; i <= y; ++i)
        dist[i] = -INF;
    Node cur, tmp;
    dist[x] = cur.dist = ;
    cur.to = x;
    q.push(cur);
    while (!q.empty())
    {
        cur = q.top(); q.pop();
        int u = cur.to;
        if (dist[u] > cur.dist)
            continue;
        for (int i = head[u]; i != -; i = g[i].next)
        {
            int v = g[i].to;
            if (dist[v] < dist[u] + g[i].dist)
            {
                tmp.dist = dist[v] = dist[u] + g[i].dist;
                tmp.to = v;
                q.push(tmp);
            }
        }
    }
    return dist[y];
}
int main()
{
    int n, Min, Max, a, b, c;
    Node tmp;
    while (scanf("%d", &n) != EOF)
    {
        e = ;
        memset(head, -, sizeof(head));
        Min = INF, Max = -INF;
        for (int i = ; i <= n; ++i)
        {
            scanf("%d%d%d", &a, &b, &c);
            a++;
            b++;
            Min = min(a - , Min);
            Max = max(b, Max);
            addEdge(a - , b, c);
        }
        for (int i = Min+; i <= Max; ++i)
        {
            addEdge(i - , i, );
            addEdge(i, i - , -);

        }
        printf("%d\n", dij(Min, Max));
    }
    return ;
}