UVALIVE 2927 "Shortest" pair of paths

裸的费用流。一开始因为这句话还觉得要拆点 样例行不通不知道这句话干啥用的。Further, the company cannot place the two chemicals in same depot (for any length of time) without special storage handling

一个点只能用一次？？

忽略这句话就直接费用流 此题类似dijkstra,dijkstra那道

#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 140
const int INF = 0x3f3f3f3f ;
int N,M;
queue<int>q;
struct node
{
    int u,v,next;
    int flow,cap,cost;
}edge[MAXN * MAXN * 4];
int cnt,src,tag;
int C,F;
bool inq[MAXN];int d[MAXN];
int head[MAXN],p[MAXN];
void add(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++;
    //反向
    edge[cnt].v = u;
    edge[cnt].u = v;
    edge[cnt].flow = 0;
    edge[cnt].cap = 0;
    edge[cnt].cost = - cost;
    edge[cnt].next = head[v];
    head[v] = cnt++;
}
void read()
{
    cnt = 0;
    memset(head,-1,sizeof(head));
    src = 0; tag = N + 1;
    add(src,1,2,0);
    //for (int i = 1; i <= N; i++)  add(i,i + N,1,0);
    for (int i = 1; i <= M; i++)
    {
        int u ,v ,w;
        scanf("%d%d%d",&u,&v,&w);
        u++;v++;
        //u的后向点 链接 v 的前向点
        add(u,v,1,w);
    }
    add(N,tag,2,0);
}
bool SPFA(int s, int t)
{
    while (!q.empty()) q.pop();
    memset(inq,false,sizeof(inq));
    memset(d,0x3f,sizeof(d));
    memset(p,-1,sizeof(p));
    d[s] = 0;
    q.push(s);
    inq[s] = true;
    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 (d[v] > d[u] + edge[i].cost && edge[i].cap > edge[i].flow)
            {
                d[v] = d[u] + edge[i].cost;
                p[v] = i;
                if (!inq[v])
                {
                    q.push(v);
                    inq[v] = true;
                }
            }
        }
    }
    return d[tag] != INF;
}
void slove()
{
    C = F = 0;
    while(SPFA(src,tag))
    {
        int a = INF;
        for (int i = p[tag]; i != -1; i = p[edge[i].u])
            a = min(a,edge[i].cap - edge[i].flow);
        for (int i = p[tag]; i != -1; i = p[edge[i].u])
        {
            edge[i].flow += a;
            edge[i ^ 1].flow -= a;
        }
        C += d[tag] * a;
        F += a;
    }
}
int main()
{
    //freopen("sample.txt","r",stdin);
    int kase = 1;
    while (scanf("%d%d",&N,&M) != EOF)
    {
        if (N == 0 && M == 0) break;
        read();
        slove();
        if (F < 2) printf("Instance #%d: Not possible
",kase++);
        else printf("Instance #%d: %d
",kase++,C);
    }
    return 0;
}