2016ACM/ICPC亚洲区青岛站 G

A coding contest will be held in this university, in a huge playground. The whole playground would be divided into N blocks, and there would be M directed paths linking these blocks. The i-th path goes from the uiui-th block to the vivi-th block. Your task is to solve the lunch issue. According to the arrangement, there are sisi competitors in the i-th block. Limited to the size of table, bibi bags of lunch including breads, sausages and milk would be put in the i-th block. As a result, some competitors need to move to another block to access lunch. However, the playground is temporary, as a result there would be so many wires on the path. 
For the i-th path, the wires have been stabilized at first and the first competitor who walker through it would not break the wires. Since then, however, when a person go through the i - th path, there is a chance of pipi to touch 
the wires and affect the whole networks. Moreover, to protect these wires, no more than cici competitors are allowed to walk through the i-th path. 
Now you need to find a way for all competitors to get their lunch, and minimize the possibility of network crashing. 

Input

The first line of input contains an integer t which is the number of test cases. Then t test cases follow. 
For each test case, the first line consists of two integers N (N ≤ 100) and M (M ≤ 5000). Each of the next N lines contains two integers si and bibi (sisi , bibi ≤ 200). 
Each of the next M lines contains three integers uiui , vivi and ci(cici(ci ≤ 100) and a float-point number pipi(0 < pipi < 1). 
It is guaranteed that there is at least one way to let every competitor has lunch.

Output

For each turn of each case, output the minimum possibility that the networks would break down. Round it to 2 digits.

Sample Input

14 42 00 33 00 31 2 5 0.53 2 5 0.51 4 5 0.53 4 5 0.5

Sample Output

0.50

题意：以桌子为节点，每张桌子上有s个选手，b袋食物，要让所有的选手都能吃上食物，然后选手在桌子连成的边 上有容量，而且除了第一次路过以外，有pi的概率破坏现场的网线（跟字面意思）。求所有选手都吃上饭且最小破坏线路的概率。

处理初始节点的时候，建立一个源点跟汇点，让 s - b > 0及选手没饭吃的连源点， s - b < 0 桌子上有饭的连汇点。

概率求我们可以 1 - （最大不破坏线路的概率） = 最小破坏线路的概率 ，用概率来代表花费。

因为求概率是乘，最短（长）路的松弛是加减。。所以取log后再跑最短路，然后最后再开方就好了。

#include <cstdio>#include <cstring>#include <iostream>#include <queue>#include <algorithm>#include <vector>#include <cmath>#define INF 1e9using namespace std;const int maxn = 200 + 10;const double esp = 1e-8;struct Edge {    int from, to, cap, flow;    double cost;    Edge(){}    Edge(int f, int t, int c, int fl, double co) : from(f), to(t), cap(c), flow(fl), cost(co){}};struct MCMF {    int n,m,s,t;    vector<Edge> edges;    vector<int> G[maxn];    bool inq[maxn];     //是否在队列    double d[maxn];        //Bellman_ford单源最短路径    int p[maxn];        //p[i]表从s到i的最小费用路径上的最后一条弧编号    int a[maxn];        //a[i]表示从s到i的最小残量            //初始化    void init(int n,int s,int t) {        this->n = n, this->s = s, this->t = t;        edges.clear();        for (int i = 0; i < n; ++i)            G[i].clear();    }            //添加一条有向边    void AddEdge(int from, int to, int cap, double cost) {        edges.push_back(Edge(from, to, cap, 0, cost));        edges.push_back(Edge(to, from, 0, 0, -cost));        m = edges.size();        G[from].push_back(m - 2);        G[to].push_back(m - 1);    }            //求一次增广路    bool BellmanFord(int &flow, double &cost) {        for (int i = 0; i < n; ++i)            d[i] = INF;        memset(inq, 0, sizeof(inq));        d[s] = 0, a[s] = INF, inq[s] = true, p[s] = 0;        queue<int> Q;        Q.push(s);        while (!Q.empty()) {            int u = Q.front();            Q.pop();            inq[u] = false;            for (int i = 0; i < G[u].size(); ++i) {                Edge &e = edges[G[u][i]];                if (e.cap > e.flow && d[e.to] - d[u] - e.cost > esp) {                    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]) {                        Q.push(e.to);                        inq[e.to] = true;                    }                }            }        }        if (d[t] == INF)            return false;        flow += a[t];        cost += a[t] * d[t];        int u = t;        while (u != s) {            edges[p[u]].flow += a[t];            edges[p[u] ^ 1].flow -= a[t];            u = edges[p[u]].from;        }        return true;    }            //求出最小费用最大流    double Min_cost() {        int flow = 0;        double cost = 0.0;        while (BellmanFord(flow, cost));        return cost;    }}MM;int main() {        //freopen("in.txt", "r", stdin);    int t;    scanf("%d", &t);    while (t--) {        int n, m;        scanf("%d%d", &n, &m);        MM.init(n + 5, 0, n + 1);        for (int i = 1; i <= n; ++i) {            int s, b, temp;            scanf("%d%d", &s, &b);            temp = s - b;            if (temp > 0) {                MM.AddEdge(0, i, temp, 0);            } else if (temp < 0) {                MM.AddEdge(i, n + 1, -temp, 0);            }        }        for (int i = 0; i < m; ++i) {            int u, v, w;            double cost;            scanf("%d%d%d%lf", &u, &v, &w, &cost);            if (w - 1 > 0)                MM.AddEdge(u, v, w - 1, -log2(1 - cost));            if (w > 0)                MM.AddEdge(u, v, 1, 0);        }        double c;        c = MM.Min_cost();        printf("%.2lf\n", 1.0 - pow(2, -c));    }}