UVa 10806 - Dijkstra, Dijkstra. (费用流)
题意
求有权无向图中来回的最短路,第一次走过的路不可以再走。
想不通为什么不能求两次最短路。这题竟然是网络流的模型,涨姿势了。
把所有边的容量设为1,建一个源点,到点1的容量2,建一个汇点,到点n的容量2。然后跑最小费用最大流,当流量达到2的时候说明可以,输出cost。
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 <ctime>
#include <cstdlib>
#include <fstream>
#include <string>
#include <sstream>
#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 = 110 + 5;
const int MOD = 20071027;
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 n, m, st, ed;
vector<EDGE> edges;
vector<int> G[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->n = n;
for (int i = 0; i < n; i++) G[i].clear();
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);
G[from].PB(m - 2);
G[to].PB(m - 1);
}
bool SPFA(int st, int ed, int &flow, int &cost)
{
fill(d, d + n + 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 = 0; i < SZ(G[u]); i++)
{
EDGE &e = edges[G[u][i]];
if (e.cap > e.flow && d[e.to] > d[u] + e.cost)
{
d[e.to] = d[u] + e.cost;
p[e.to] = G[u][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;
}
void min_cost(int st, int ed)
{
int flow = 0, cost = 0;
while (SPFA(st, ed, flow, cost))
{
if (flow == 2)
{
printf("%d\n", cost);
return;
}
}
puts("Back to jail");
}
}MC;
int main()
{
//ROP;
int n, m;
while (scanf("%d", &n), n)
{
scanf("%d", &m);
MC.init(n + 5);
MC.add_edge(0, 1, 2, 0);
MC.add_edge(n, n + 1, 2, 0);
while (m--)
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
MC.add_edge(a, b, 1, c);
MC.add_edge(b, a, 1, c);
}
MC.min_cost(0, n + 1);
}
return 0;
}