UVALive

题意:每个点i有(s_i)个人和(b_i)份食物,每个人都要找到一份食物.现在有M条有向边,从点i到点j,容量为c,第一次走过不要紧,从第二次开始就要承担(p(0<p<1))的道路损坏的风险.题目保证每个人都能拿到食物,求这个风险的最小值.
分析:建立源点S和汇点T.从S到点i建容量为(s_i),费用为0的边;从点i到T建容量为(b_i),费用为0的边.
风险的最小值可以转化为求安全达成目标的. 则(ans = prod (1-p_i)^{k_i}),
对ans取对数,(log(ans) = sum k_i*log(1-p_i)).因为需要求这个值最大,所以费用需要取反.
因为第一次走没有风险,可以将这条容量为1的边分离出来,若容量还有剩余则建容量为(c-1),费用为(-log(1-p))的边.
跑一遍费用流,费用取反再还原之后就是最大的安全通过的概率,1减去这个概率就是最小的风险.

#include<bits/stdc++.h>
using namespace std;
#define eps 1e-7
const int MAXN = 10005;
const int MAXM = 100005;
const int INF = 0x3f3f3f3f;
struct Edge{
    int to, next, cap, flow;
    double cost;
} edge[MAXM];
int head[MAXN], tot;
int pre[MAXN];
double dis[MAXN];
bool vis[MAXN];
int N; 
void init(int n)
{
    N = n;
    tot = 0;
    memset(head, -1, sizeof(head));
}

void AddEdge(int u, int v, int cap, double cost)
{
    edge[tot] = (Edge){v,head[u],cap,0,cost};
    head[u] = tot++;
    edge[tot] = (Edge){u,head[v],0,0,-cost};
    head[v] = tot++;
}

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

            if (edge[i].cap > edge[i].flow && dis[v] - (dis[u] + edge[i].cost)> eps){
                dis[v] = dis[u] + edge[i].cost;
                pre[v] = i;
                if (!vis[v]){
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
    if (pre[t] == -1) return false;
    else  return true;
}

int minCostMaxflow(int s, int t, double &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].to]){
            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].to]){
            edge[i].flow += Min;
            edge[i ^ 1].flow -= Min;
            cost += edge[i].cost * Min;
        }
        flow += Min;
    }
    return flow;
}

int have[MAXN];
int need[MAXN];


int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
    #endif
    int T;
    scanf("%d",&T);
    while(T--){
        int n,m;
        scanf("%d %d",&n,&m);
        init(n+2);
        int u,v; double w;
        int s= 0, t = n+1;
        for(int i=1 ; i<=n; ++i){
            scanf("%d %d", &need[i], &have[i]);
        }
        for(int i=1;i<=n;++i){
            AddEdge(s,i,need[i],0);
            AddEdge(i,t,have[i],0);
        }

        int c;
        double p;
        for(int i=1;i<=m;++i){
            scanf("%d %d %d %lf",&u, &v, &c, &p);
            if(c >= 1) AddEdge(u,v,1,0.0);
            if(c > 1) AddEdge(u,v,c-1,-log2(1.0-p));
        }
        double cost = 0.0;
        minCostMaxflow(s,t,cost);
        //cout<<cost<<endl;
        printf("%.2f
",1.0-pow(2.0,-cost));
    }
    return 0;
}