POJ1201 Intervals 【差分约束】

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
",i,d[i]);*/
	printf("%d
",d[N - 1]);
	return 0;
}