[BZOJ2330][SCOI2011][拓扑排序][强连通分量][Tarjan]Candy

[Problem Description]
幼儿园里有N个小朋友，lxhgww老师现在想要给这些小朋友们分配糖果，要求每个小朋友都要分到糖果。但是小朋友们也有嫉妒心，总是会提出一些要求，比如小明不希望小红分到的糖果比他的多，于是在分配糖果的时候，lxhgww需要满足小朋友们的K个要求。幼儿园的糖果总是有限的，lxhgww想知道他至少需要准备多少个糖果，才能使得每个小朋友都能够分到糖果，并且满足小朋友们所有的要求。
[Algorithm]
强连通分量tarjan，拓扑排序
[Analysis]
这是经典的拓扑排序的模型。蛋疼的是加入了可以等于和等于小于的关系。没关系，只要解决了等于的问题，照样可以拓扑排序做。先将等于和等于小于的边加入图中（等于为双向边，等于小于为单向边），然后跑强连通分量，这样每一个强连通分量一定糖果数相等，且原图变成了一个有向无环图（如果数据有解的话），跑拓扑排序即可。
[Pay Attention]
最后的结果要用long long，还有就是跑完强连通分量要注意检查一下，可能存在强连通分量内部出现大于或小于的边，这样的话就是无解

[code]/**************************************************************
Problem: 2330
User: gaotianyu1350
Language: C++
Result: Accepted
Time:352 ms
Memory:14156 kb
****************************************************************/

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <iostream>
#include <queue>
using namespace std;

#define MAXN 201000

struct relation
{
int Type;
int x, y;
};

relation q[MAXN];
int cct[MAXN] = {0}, size[MAXN] = {0}, candy[MAXN], cnt = 0;
long long f[MAXN] = {0};
int rudu[MAXN] = {0};
int point[MAXN], nxt[MAXN], v[MAXN];
int dfsTime[MAXN] = {0}, lessTime[MAXN] = {0}, st[MAXN], top = 0;
int nowTime = 0;
bool same[MAXN];
int tot, n, k;

inline void swap(int &a, int &b)
{
int temp = a; a = b; b = temp;
}

inline void clear()
{
memset(point, 0, sizeof(point));
memset(nxt, 0, sizeof(nxt));
memset(rudu, 0, sizeof(rudu));
memset(same, 0, sizeof(same));
tot = 0;
}

inline void addedge(int x, int y, bool canbesame)
{
tot++;
nxt[tot] = point[x];
point[x] = tot;
v[tot] = y;
same[tot] = canbesame;
rudu[y]++;
}

void tarjan(int now)
{
dfsTime[now] = lessTime[now] = ++nowTime;
st[++top] = now;
for (int temp = point[now]; temp; temp = nxt[temp])
{
int tar = v[temp];
if (!dfsTime[tar])
{
tarjan(tar);
lessTime[now] = min(lessTime[now], lessTime[tar]);
}
else
if (!cct[tar])
lessTime[now] = min(lessTime[now], dfsTime[tar]);
}
if (lessTime[now] == dfsTime[now])
{
cnt++;
while (1)
{
cct[st[top--]] = cnt;
size[cnt]++;
if (st[top + 1] == now) break;
}
}
}

long long Solve()
{
queue<int> q;
long long ans = 0;
int already = 0;
for (int i = 1; i <= cnt; i++)
if (!rudu[i])
{
f[i] = 1;
q.push(i);
}
if (q.empty()) return -1;
while (!q.empty())
{
int now = q.front(); q.pop();
ans += size[now] * f[now];
already += size[now];
for (int temp = point[now]; temp; temp = nxt[temp])
{
int tar = v[temp];
if (same[temp])
f[tar] = f[tar] > f[now] ? f[tar] : f[now];
else
f[tar] = f[tar] > f[now] ? f[tar] : f[now] + 1;
rudu[tar]--;
if (!rudu[tar])
q.push(tar);
}
}
if (already < n) return -1;
else return ans;
}

int main()
{
//freopen("input.txt", "r", stdin);
clear();
scanf("%d%d", &n, &k);
for (int i = 1; i <= k; i++)
{
scanf("%d%d%d", &q[i].Type, &q[i].x, &q[i].y);
if (q[i].Type == 3 || q[i].Type == 4)
swap(q[i].x, q[i].y);
if (q[i].Type == 1)
addedge(q[i].x, q[i].y, false), addedge(q[i].y, q[i].x, false);
if (q[i].Type == 3 || q[i].Type == 5)
addedge(q[i].x, q[i].y, false);
}
for (int i = 1; i <= n; i++)
if (!dfsTime[i])
tarjan(i);
clear();
for (int i = 1; i <= k; i++)
if (cct[q[i].x] != cct[q[i].y])
{
if (q[i].Type == 3 || q[i].Type == 5)
addedge(cct[q[i].x], cct[q[i].y], true);
else
addedge(cct[q[i].x], cct[q[i].y], false);
}
else
{
if (q[i].Type == 2 || q[i].Type == 4)
{
printf("-1\n");
return 0;
}
}
long long ans = Solve();
printf("%lld\n", ans);
}