题目链接

POJ1201

题解

差分约束

令\(a[i]\)表示是否选择\(i\)，\(s[i]\)表示\(a[i]\)的前缀和

对\(s[i] \quad i \in [-1,50000]\)分别建立一个点

首先有

\[s[i] - s[i - 1] \ge 0\]

\[s[i] - s[i - 1] \le 1\]

然后就是限制条件

\[s[b] - s[a - 1] \ge c\]

然后就没了

用\(spfa\)跑最长路

由于题目保证有解，所以不会存在正环

复杂度上界是\(O(nm)\)的，但由于保证有解，而且\(spfa\)的玄学复杂度，并不会\(T\)

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,-0x3f3f3f3f,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 50005,maxm = 200005,INF = 1000000000;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
int h[maxn],ne,N = 50001;
struct EDGE{int to,nxt,w;}ed[maxm];
inline void build(int u,int v,int w){
    ed[++ne] = (EDGE){v,h[u],w}; h[u] = ne;
}
queue<int> q;
int d[maxn],vis[maxn];
void spfa(){
    for (int i = 0; i <= N; i++) d[i] = -INF; d[N] = 0;
    q.push(N);
    int u;
    while (!q.empty()){
        u = q.front(); q.pop();
        vis[u] = false;
        Redge(u) if (d[to = ed[k].to] < d[u] + ed[k].w){
            d[to] = d[u] + ed[k].w;
            if (!vis[to]) q.push(to),vis[to] = true;
        }
    }
}
int main(){
    int m = read(),a,b,c;
    while (m--){
        a = read(); b = read(); c = read();
        a--; if (a == -1) a = N;
        build(a,b,c);
    }
    build(N,0,0); build(0,N,-1);
    for (int i = 1; i < N; i++)
        build(i - 1,i,0),build(i,i - 1,-1);
    spfa();
    /*for (int i = 0; i < 15; i++)
        printf("d[%d] = %d\n",i,d[i]);*/
    printf("%d\n",d[N - 1]);
    return 0;
}

原文地址：https://www.cnblogs.com/Mychael/p/9160278.html