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

然后就是限制条件

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;

}