UVA 1658 (费用流)

求两条不相交从1到n路径的长度和的最小值.

每个点拆成入点出点,流量1费用为边权,拆成的两个点流量为1费用为0,除了源点和汇点流量为2.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
#define INF 11111111
#define maxn 2111
#define maxm 2111111

int n, m;
int s, t;
struct node {
int u, v, next, cap, flow, cost;
}edge[maxm];
int head[maxn], cnt;
int pre[maxn], dis[maxn];
bool vis[maxn];
int N;

void init () {
memset (head, -1, sizeof head);
cnt = 0;
}

void add_edge (int u, int v, int cap, int cost) {
edge[cnt].u = u, edge[cnt].v = v, edge[cnt].cap = cap, edge[cnt].flow = 0;
edge[cnt].cost = cost, edge[cnt].next = head[u], head[u] = cnt++;
edge[cnt].u = v, edge[cnt].v = u, edge[cnt].cap = 0, edge[cnt].flow = 0;
edge[cnt].cost = -cost, edge[cnt].next = head[v], head[v] = cnt++;
}

bool spfa (int s, int t) {
queue <int> q;
for (int i = 0; i < N; i++) {
dis[i] = INF;
vis[i] = 0;
pre[i] = -1;
}
dis[s] = 0;
vis[s] = 1;
q.push (s);
while (!q.empty ()) {
int u = q.front (); q.pop ();
vis[u] = 0;
for (int i = head[u]; i != -1; i = edge[i].next) {
int v = edge[i].v;
if (edge[i].cap > edge[i].flow && dis[v] > dis[u]+edge[i].cost) {
dis[v] = dis[u]+edge[i].cost;
pre[v] = i;
if (!vis[v]) {
vis[v] = 1;
q.push (v);
}
}
}
}
if (pre[t] == -1)
return 0;
else
return 1;
}

int MCMF (int s, int t, int &cost) {
int flow = 0;
cost = 0;
while (spfa (s, t)) {
int Min = INF;
for (int i = pre[t]; i != -1; i = pre[edge[i^1].v]) {
if (Min > edge[i].cap-edge[i].flow) {
Min = edge[i].cap-edge[i].flow;
}
}
for (int i = pre[t]; i != -1; i = pre[edge[i^1].v]) {
edge[i].flow += Min;
edge[i^1].flow -= Min;
cost += edge[i].cost*Min;
}
flow += Min;
}
return flow;
}

int main () {
//freopen ("in.txt", "r", stdin);
while (scanf ("%d%d", &n, &m) == 2) {
init ();
add_edge (0, 1, 2, 0);
add_edge (2*(n-1), 2*(n-1)+1, 2, 0);
for (int i = 1; i < n-1; i++) {
add_edge (i*2, i*2+1, 1, 0);
}
for (int i = 0; i < m; i++) {
int u, v, w;
scanf ("%d%d%d", &u, &v, &w);
u--, v--;
add_edge (2*u+1, 2*v, 1, w);
}
int cost = 0;
N = 2*n;
s = 0, t = 2*(n-1);
MCMF (s, t, cost);
printf ("%d\n", cost);
}
return 0;
}