洛谷 #3275. 糖果

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题意

opt=1，表示第A个小朋友分到的糖果必须和第B个小朋友分到的糖果一样多

opt=2，表示第A个小朋友分到的糖果必须少于第B个小朋友分到的糖果

opt=3，表示第A个小朋友分到的糖果必须不少于第B个小朋友分到的糖果

opt=4，表示第A个小朋友分到的糖果必须多于第B个小朋友分到的糖果

opt=5，表示第A个小朋友分到的糖果必须不多于第B个小朋友分到的糖果

每个小朋友的糖果数>=1，问最小需要糖果数

题解

差分约束

opt=1：A到B连0边，B到A连0边

opt=2：A到B连1边

opt=3：B到A连0边

opt=1：B到A连1边

opt=1：A到B连0边

然后SPFA跑最长路

调试记录

建正权边要跑最长路

#include <cstdio>
#include <cstring>
#include <queue>
#include <cctype>
#define maxn 100005
#define int long long

using namespace std;

struct node{
	int to, next, l;
}e[maxn << 2];
int head[maxn], tot = 0;
void addedge(int u, int v, int l){
	e[++tot].to = v, e[tot].next = head[u], e[tot].l = l;
	head[u] = tot;
}

int n, k, dis[maxn], t[maxn];
bool inq[maxn];
bool SPFA(int s){
	dis[s] = 0; inq[s] = true;
	queue <int> q; while (!q.empty()) q.pop();
	q.push(s);
	while (!q.empty()){
		int cur = q.front(); q.pop(); inq[cur] = false;
		for (int i = head[cur]; i; i = e[i].next){
			if (dis[e[i].to] < dis[cur] + e[i].l){
				dis[e[i].to] = dis[cur] + e[i].l;
				if (!inq[e[i].to]){
					if (++t[e[i].to] >= n) return false;
					q.push(e[i].to);
					inq[e[i].to] = true;
				}
			}
		}
	}
	return true;
}

inline void read(int &x){
	x = 0; char ch = getchar();
	while (!isdigit(ch)) ch = getchar();
	while (isdigit(ch)) x = (x * 10) + (ch & 15), ch = getchar();
}

signed main(){
	memset(dis, -0x3f, sizeof dis);
	read(n), read(k);
	for (int u, v, opt, i = 1; i <= k; i++){
		read(opt), read(u), read(v);
		if (opt == 1) addedge(u, v, 0), addedge(v, u, 0);
		if (opt == 2){
			if (u == v){
				puts("-1");
				return 0;
			}
			addedge(u, v, 1);
		}
		if (opt == 3) addedge(v, u, 0);
		if (opt == 4){
			if (u == v){
				puts("-1");
				return 0;
			}
			addedge(v, u, 1);
		}
		if (opt == 5) addedge(u, v, 0);
	}
	for (int i = n; i >= 1; i--) addedge(0, i, 1);
	if (!SPFA(0)) puts("-1");
	else{
		int res = 0;
		for (int i = 1; i <= n; i++) res += dis[i];
		printf("%d\n", res);
	}
	
	return 0;
}