HDU 1853 - Cyclic Tour (最小费用最大流)
题意
求把每个点都属于一个环,环的最小权和。
思路
学习了最小费用最大流求环的最小权的方法。
把一个点拆成两个,控制入度和出度,如果最后的流量等于一开始的流量,就可以。
不过奇怪的是在另外一题完全一样的题里我的模板竟然TLE了。。。再多做几题看看
代码
#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
#define LL long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 300 + 10;
const int MOD = 1e9 + 7;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
struct EDGE
{
int from, to, cap, flow, cost;
};
struct MCMF
{
int m, st, ed;
vector<EDGE> edges;
int head[MAXN], next[MAXN * MAXN];
int inq[MAXN]; //是否在队列中
int d[MAXN]; //bellman-Ford
int p[MAXN]; //上一条弧
int a[MAXN]; //可改进量
void init(int n)
{
this->ed = n;
MS(head, -1);
edges.clear();
}
void add_edge(int from, int to, int cap, int cost)
{
edges.PB((EDGE){from, to, cap, 0, cost});
edges.PB((EDGE){to, from, 0, 0, -cost});
m = SZ(edges);
next[m - 2] = head[from]; head[from] = m - 2;
next[m - 1] = head[to]; head[to] = m - 1;
}
bool SPFA(int st, int ed, int &flow, int &cost)
{
fill(d, d + ed + 1, INF);
MS(inq, 0);
d[st] = 0; inq[st] = 1; p[st] = -1; a[st] = INF;
queue<int> qu;
qu.push(st);
while (!qu.empty())
{
int u = qu.front(); qu.pop();
inq[u] = 0;
for (int i = head[u]; i != -1; i = next[i])
{
EDGE &e = edges[i];
if (e.cap > e.flow && d[e.to] > d[u] + e.cost)
{
d[e.to] = d[u] + e.cost;
p[e.to] = i;
a[e.to] = min(a[u], e.cap - e.flow);
if (!inq[e.to])
qu.push(e.to), inq[e.to] = 1;
}
}
}
if (d[ed] == INF) return false;
flow += a[ed]; cost += d[ed] * a[ed];
int u = ed;
while (u != st)
{
edges[p[u]].flow += a[ed];
edges[p[u] ^ 1].flow -= a[ed];
u = edges[p[u]].from;
}
return true;
}
}MFMC;
int main()
{
//ROP;
int T, i, j, ncity, nway;
while (~scanf("%d%d", &ncity, &nway))
{
int st = 0, ed = 2 * ncity + 1;
MFMC.init(ed);
while (nway--)
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
MFMC.add_edge(a, b + ncity, 1, c);
}
for (i = 1; i <= ncity; i++)
{
MFMC.add_edge(st, i, 1, 0);
MFMC.add_edge(i + ncity, ed, 1, 0);
//MFMC.add_edge(i, i + ncity, 1, 0);
}
int cost = 0, flow = 0;
while (MFMC.SPFA(st, ed, flow, cost));
if (flow == ncity) printf("%d\n", cost);
else printf("-1\n");
}
return 0;
}