UVA 10806 Dijkstra, Dijkstra.

费用流第一题主要是临街表实现这个算法的问题。这里存下

思路还是比较简单。源点0，汇点N+1.费用为边长。容量为1.（普通边）。添加边为2（0-1 N-N+1）

代码

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);}
#define MAXN 110
const int INF = 0x3f3f3f3f ;
struct node
{
    int u,v,next;
    int flow,cost;
    int cap;
}edge[20100];
int head[MAXN],d[MAXN];
bool inq[MAXN];
int p[MAXN],cnt;
int c,ans,N,M;
void add(int u, int v, int cost, int cap)
{
    edge[cnt].v = v;
    edge[cnt].u = u;
    edge[cnt].cost = cost;
    edge[cnt].cap = cap;
    edge[cnt].flow = 0;
    edge[cnt].next = head[u];
    head[u] = cnt++;
    // 反向边
    edge[cnt].v = u;
    edge[cnt].u = v;
    edge[cnt].cost = -cost;
    edge[cnt].cap = 0;
    edge[cnt].flow = 0;
    edge[cnt].next = head[v];
    head[v] = cnt++;
}
bool spfa(int s,int t)
{
    queue<int>q;
    while (!q.empty()) q.pop();
    q.push(s);
    memset(inq,false,sizeof(inq));
    inq[s] = true;
    memset(d,0x3f,sizeof(d));
    d[0] = 0;
    memset(p,-1,sizeof(p));
    while (!q.empty())
    {
        int u = q.front();
        q.pop();
        inq[u] = false;
        for (int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].v;
            if (edge[i].cap > edge[i].flow && d[v] > d[u] + edge[i].cost)
            {
                d[v] = d[u] + edge[i].cost;
                p[v] = i;
                if (!inq[v])
                {
                    q.push(v);
                    inq[v] = true;
                }
            }
        }
    }
    return d[N + 1] != INF;
}
void slove(int s,int t)
{
    c = 0 ,ans = 0;
    while (spfa(s,t))
    {
        int a = INF;
        for (int i = p[N + 1]; i != -1; i = p[edge[i].u]) a = min(a,edge[i].cap - edge[i].flow);
        for (int i = p[N + 1]; i != -1; i = p[edge[i].u])
        {
            edge[i].flow += a;
            edge[i ^ 1].flow -= a;
        }
        c += d[N + 1] * a;
        ans += a;
    }
}
int main()
{
    //freopen("sample.txt","r",stdin);
    while (scanf("%d",&N)!=EOF)
    {
        if (N == 0) break;
        memset(head,-1,sizeof(head));
        cnt = 0;
        scanf("%d",&M);
        while(M--)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w,1);
            add(v,u,w,1);
        }
        add(0,1,0,2);
        add(N,N + 1,0,2);
        slove(0,N + 1);
        if (ans >= 2) printf("%d
",c);
        else puts("Back to jail");
    }
    return 0;
}