[BZOJ-2330]糖果

Description

有n个人,每个人会被分到一些糖果。
有m对关于两个人分到的糖果个数的限制关系。
求所有人获得糖果总数的最小值。

Solution

这是一道[SCOI2011]的题。
用差分约束系统做,跑一遍最长路就可以了。

Notice

对0加边的时候,如果正序加边的话好像会TLE,所以要倒序加边。
注意要判断无解的情况。

Code

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#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define sqz main
#define ll long long
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
#define per(i, a, b) for (int i = (a); i >= (b); i--)
#define Rep(i, a, b) for (int i = (a); i < (b); i++)
#define travel(i, u) for (int i = head[u]; ~i; i = edge[i].next)

const ll INF = 1e9, Mo = 998244353;
const int N = 100001;
const double eps = 1e-6;
namespace slow_IO
{
ll read()
{
ll x = 0; int zf = 1; char ch = getchar();
while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
return x * zf;
}
void write(ll y)
{
if (y < 0) putchar('-'), y = -y;
if (y > 9) write(y / 10);
putchar(y % 10 + '0');
}
}
using namespace slow_IO;

int head[N + 5], Dis[N + 5], Flag[N + 5], Q[N + 5], Vis[N + 5];
int edgenum = 0, n, k;
ll ans;
struct node
{
int vet, next, val;
}edge[3 * N + 5];
void addedge(int u, int v, int w)
{
edge[edgenum].vet = v;
edge[edgenum].val = w;
edge[edgenum].next = head[u];
head[u] = edgenum++;
}

int spfa()
{
int Head = 0, Tail = 1;
Q[Head] = 0; Flag[0] = 1; Vis[0] = 1;
while (Head != Tail)
{
int u = Q[Head]; Head = (Head + 1) % N; Flag[u] = 0;
travel(i, u)
{
int v = edge[i].vet;
if (Dis[u] + edge[i].val > Dis[v])
{
Dis[v] = Dis[u] + edge[i].val;
if (++Vis[v] >= n) return 0;
if (!Flag[v])
{
Q[Tail] = v; Flag[v] = 1;
Tail = (Tail + 1) % N;
}
}
}
}
return 1;
}

int sqz()
{
n = read(), k = read(), ans = 0;
rep(i, 0, n) head[i] = -1;
rep(i, 1, k)
{
int op = read(), x = read(), y = read();
if ((op == 2 || op == 4) && x == y)
{
puts("-1");
return 0;
}
if (op == 1) addedge(x, y, 0), addedge(y, x, 0);
if (op == 2) addedge(x, y, 1);
if (op == 3) addedge(y, x, 0);
if (op == 4) addedge(y, x, 1);
if (op == 5) addedge(x, y, 0);
}
per(i, n, 1) addedge(0, i, 1);
if (!spfa())
{
puts("-1");
return 0;
}
rep(i, 1, n) ans += Dis[i];
printf("%lld\n", ans);
return 0;
}
文章目录
  1. 1. Description
  2. 2. Solution
  3. 3. Notice
  4. 4. Code