【差分约束】 HDOJ 1534 Schedule Problem

不难的差分约束。。列出四类不等式，建好图就可以了。。。

#include <iostream>  
#include <queue>  
#include <stack>  
#include <map>  
#include <set>  
#include <bitset>  
#include <cstdio>  
#include <algorithm>  
#include <cstring>  
#include <climits>  
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 100005
#define maxm 300005
#define eps 1e-10
#define mod 998244353
#define INF 999999999
#define lowbit(x) (x&(-x))
#define mp mark_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid  
#define rson o<<1 | 1, mid+1, R  
typedef long long LL;
//typedef int LL;
using namespace std;
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
// head

int inq[maxn], outq[maxn], dis[maxn], t[maxn];
int h[maxn], next[maxm], v[maxm], cost[maxm];
queue<int> q;
char ss[maxn];
int cnt, n, m;

void init(void)
{
	memset(h, -1, sizeof h);
	memset(inq, 0, sizeof inq);
	memset(outq, 0, sizeof outq);
	for(int i = 0; i <= n; i++) dis[i] = -INF;
}
void read(void)
{
	for(int i = 1; i <= n; i++) scanf(t[i]);
}
bool spfa(int s)
{
	while(!q.empty()) q.pop();
	dis[s] = 0, q.push(s), inq[s] = 1;
	while(!q.empty()) {
		int u = q.front(); q.pop(), inq[u] = 0, outq[u]++;
		if(outq[u] > n+1) return false;
		for(int e = h[u]; ~e; e = next[e])
			if(dis[u] + cost[e] > dis[v[e]]) {
				dis[v[e]] = dis[u] + cost[e];
				if(!inq[v[e]]) q.push(v[e]), inq[v[e]] = 1;
			}
	}
	return true;
}
void addedges(int u, int vv, int c)
{
	next[cnt] = h[u], h[u] = cnt, v[cnt] = vv, cost[cnt] = c, cnt++;
}
void work(void)
{
	int a, b;
	while(scanf("%s", ss), ss[0] != '#') {
		//scanf("%d%d", &a, &b);
		scanf(a), scanf(b);
		if(ss[0] == 'S' && ss[2] == 'S') addedges(b, a, 0);
		if(ss[0] == 'S' && ss[2] == 'F') addedges(b, a, t[b]);
		if(ss[0] == 'F' && ss[2] == 'S') addedges(b, a, -t[a]);
		if(ss[0] == 'F' && ss[2] == 'F') addedges(b, a, t[b] - t[a]);
	}
	for(int i = 1; i <= n; i++) addedges(0, i, 0);
}
int main(void)
{
	int _ = 0;
	while(scanf("%d", &n), n != 0) {
		read();
		init();
		work();
		printf("Case %d:\n", ++_);
		if(!spfa(0)) printf("impossible\n");
		else for(int i = 1; i <= n; i++) printf("%d %d\n", i, dis[i]);
		printf("\n");
	}
	return 0;
}