POJ 1201 Intervals（差分约束系统）

题意：求一个包含元素最少集合Z，满足以下条件

给定区间[ai, bi]使得Z里面有 >= ci 个元素在区间[ai, bi]里面，求最少的元素个数

思路：

设 s[i]：集合Z里面 <= i的元素个数，那么由题意：s[ai] - s[bi] >= ci ，这是差分约束系统，光是这个建立不了通路，所以还有 0 <= s[i] - s[i - 1] <= 1，这样就建立起来了,由于一定有解，故不存在负环

#include
#include
#include
#include
#include
#include
#define bug printf("CCCCC\n")
const int maxn = 5 * 1e4 + 10;
const int INF = 1e9;
using namespace std;

struct P {
    int to, cost;
    P(int t, int c) : to(t), cost(c) {}
};
int d[maxn], vis[maxn], cnt[maxn];
int n, a, b, c, l, r;
vector
 G[maxn];

bool operator < (P p, P q) {
    return p.cost > q.cost;
}

void init() {
    fill(d, d + maxn, INF);
    memset(vis, 0, sizeof(vis));
    memset(cnt, 0, sizeof(cnt));
    for(int i = 0; i < maxn; i++)
        G[i].clear();
    l = INF; r = -1;
}

bool spfa(int from, int to) {
    vis[from] = 1;
    d[from] = 0;
    queue q;
    q.push(from);
    while(!q.empty()) {
        int u = q.front(); q.pop();
        vis[u] = 0;
        for(int i = 0; i < G[u].size(); i++) {
            int v = G[u][i].to;
            if(d[v] > d[u] + G[u][i].cost) {
                d[v] = d[u] + G[u][i].cost;
                if(vis[v]) continue;
                vis[v] = 1; q.push(v);
                if(++cnt[v] > n) return true;
            }
        }
    }
    return false;
}

int main() {
    while(scanf("%d", &n) != EOF) {
        init();
        for(int i = 0; i < n; i++) {
            scanf("%d %d %d", &a, &b, &c);
            a++; b++;
            G[b].push_back(P(a - 1, -c));
            l = min(l, a);
            r = max(r, b);
        }
        for(int i = l; i <= r; i++) {
            G[i - 1].push_back(P(i, 1));
            G[i].push_back(P(i - 1, 0));
        }
        spfa(r, l - 1);
        int mind = d[l - 1];
        printf("%d\n", -mind);
    }
    return 0;
}