HDU5988

HDU5988

题意：

　　有n个区域，每个区域有s个人，b份饭。现在告诉你每个区域间的有向路径，每条路有容量和损坏路径的概率。问如何走可以使得路径不被破坏的概率最小。第一个人走某条道路是百分百不会损坏道路的。

思路：

　　对于每个人，他从起点到目的地，不损坏道路的概率是（1 - p【1】*p【2】...*p【r】）。相乘不好做，对减号右边的乘法取对数，就成了相加，就可以愉快的做相加运算了，就是可以跑费用流了。这道题spfa还有加入esp=1e-8。

//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "
";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;
typedef pair<pii,int> pp3;
//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '
'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e8+7;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;


ll gcd(ll a, ll b) {return b?gcd(b,a%b):a;}
template<typename T>inline void mod_(T &A,ll MOD=mod) {A%=MOD; A+=MOD; A%=MOD;}

template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}


/*-----------------------show time----------------------*/
            const int maxn = 1e6+9;
            struct edge{
                int to,val,nxt;
                double cost;
            }gedge[maxn];
            int h[maxn],gpre[maxn];
            int gpath[maxn];
            double gdist[maxn];
            bool in[maxn];
            int gcount = 0,n,m;

            bool spfa(int s,int t){
                    //memset(gdist,inf,sizeof(gdist));
                    for(int i=0; i<=n+1; i++) {
                        gpre[i] = -1;
                        gdist[i] = 999999999.9;
                        in[i] = false;
                    }
                    gdist[s] = 0.0; in[s] = true;
                    queue<int>que;
                    que.push(s);
                    while(!que.empty()){
                        int u = que.front();
                        que.pop();    in[u] =false;
                        for(int e = h[u]; e!=-1; e= gedge[e].nxt){
                            int v = gedge[e].to;
                            double w = gedge[e].cost;
                            if(gedge[e].val > 0 && gdist[v] - gdist[u] - w > 1e-8){
                                gdist[v] = gdist[u] + w;
                                gpre[v] = u;
                                gpath[v] = e;
                                if(!in[v]){
                                    que.push(v); in[v] = true;
                                }
                            }
                        }
                    }

                    if(gpre[t] == -1) return false;
                    return true;
            }

            double MinCostFlow(int s,int t){
                double cost = 0.0;int flow = 0;
                while(spfa(s,t)){
                    double f = 999999999.99;
                    for(int u=t; u!=s;u = gpre[u]){
                        if(gedge[gpath[u]].val < f){
                            f = gedge[gpath[u]].val;
                        }
                    }
                    flow += f;
                    cost += 1.0*gdist[t] * f;
                    for(int u=t; u!=s; u = gpre[u]){
                        gedge[gpath[u]].val -= f;
                        gedge[gpath[u] ^ 1].val += f;
                    }
                }
                return cost;
            }

            void addedge(int u,int v,int val, double cost){
                gedge[gcount].to = v;
                gedge[gcount].val = val;
                gedge[gcount].cost = cost;
                gedge[gcount].nxt = h[u];
                h[u] = gcount++;

                gedge[gcount].to = u;
                gedge[gcount].val = 0;
                gedge[gcount].cost = -cost;
                gedge[gcount].nxt = h[v];
                h[v] = gcount++;
            }
int main(){
            int T;      scanf("%d", &T);
            while(T--){
                    memset(h,-1,sizeof(h)); gcount = 0;
                    scanf("%d%d", &n, &m);
                    for(int i=1; i<=n; i++){
                        int u,v;
                        scanf("%d%d", &u, &v);
                        if(u > v) addedge(0,i,u-v,0.0);
                        else if(u < v) addedge(i,n+1,v-u,0.0); 
                    }
                    for(int i=1; i<=m; i++){
                        int u,v,c;double p;
                        scanf("%d%d%d%lf", &u, &v, &c, &p);
                        if(c == 0)continue;
                        addedge(u,v,1,0.0);
                        addedge(u,v,c-1,-1.0*log(1-p));
                    }
                    double ans = -1.0*MinCostFlow(0,n+1);

                    printf("%.2f
", 1-exp(ans));
            }

            return 0;
}