Two points + greedy--hdu 5855

Topic Links:
http://acm.split.hdu.edu.cn/showproblem.php?pid=5855

Reference blog:
http://blog.csdn.net/queuelovestack/article/details/52222085

Analysis: Given n factories, each factory gives the construction time and cost, gives the M stores, each store gives the revenue and needs the factory supply to be profitable, the minimum time to obtain l profit (Profit = store income-the investment cost of the factory)

Exercises
The minimum time required, so we can first search time, because the time is 1~1000000000, so the binary search does not time out. Determine whether the maximum profit at this time meets the requirements after each search time.

Judging method:
We can calculate the maximum time that every store can make a profit, that is, the one that needs to be built the longest, so you can tell if the store is profitable in the current time, and then use a value to store the cost of the currently built factory. So the profit of the store = store revenue-The total cost of building the plant + the total cost of the plant has been built. Each time you select the most profitable store, update the already-established factory costs for the currently-not-selected store. Until the current cannot be selected, the maximum profit in the process is satisfied if it exceeds L, otherwise it is not satisfied.

AC Code:

/************************************************************************* > File name:1012.cpp > Author: Akira > Mail:qaq.febr2.qaq@gmail.com > Created time:2016 August 16 Tuesday 18:44 04 seconds ********************************** /#include <iostream> #include <cstdio> #include <cstring> # Include <string> #include <cstdlib> #include <algorithm> #include <queue> #include <stack > #include <map> #include <cmath> #include <vector> #include <set> #include <list> typed
EF Long Long LL;
typedef unsigned long long ULL;
typedef long double LD;

#define MST (A, B) memset (A,b,sizeof (a)) #define CLR (a) MST (a,0) #define SQR (a) ((a) * (a)) using namespace Std;

#define MAXN 100000 #define MAXM maxn*10 #define INF 1000000007 #define BUG cout<<88888888<<endl;
int T;
int n,m;

int tar;

struct Plants {int pay,t;//plant construction cost and time}p[233];        struct Shop {int pro; BusinessShop's profit int K; The factory int index[233] that the store needs;       Factory subscript int time;       Maximum time int cost for factory construction;     Total cost of the construction plant int finish;

The cost of the currently built factory is}s[233]; struct Edge {int v,next;}
EDGE[66666];
int cont;

int head[233];
    void Add (int u, int v) {edge[cont].v = v;
    Edge[cont].next = Head[u];
Head[u] = cont++;
    } void init () {cont = 0;
MST (head,-1);
} int svis[233];

int pvis[233];
    int judge (int t) {int Get =-inf;
    CLR (Svis);
    CLR (Pvis);            for (int i=1;i<=m;i++) s[i].finish = 0;                        Each store corresponds to a built factory clear 0 int Max;                        Record the maximum profit in a selected store int tmp = 0;
        Record total profit for (int i=1;i<=m;i++)//altogether M store {Max =-inf;
        int choose; for (int j=1;j<=m;j++)//Traverse every store, select the most profitable store {if (! SVIS[J] && s[j].time<=t)//If the store is not traversed and its maximum profit time does not exceed the current T time, then the judgment {I F (max<s[j].pro-s[J].cost+s[j].finish]//If the current store profit is greater than the previous selection, update the selected store {Max = S[j].pro-s[j].cos       T+s[j].finish;                                 Update profit choose = J;      Update store Subscript}} if (Max = =-inf) break;           Description No store can be selected, then jump out svis[choose] = 1;                   Mark the selected store Tmp+=max;        Update total Profit get = MAX (tmp, get); Maximum profit achieved during the update for (int j=1;j<=s[choose]. k;j++)//traverse all the factories required by the current selection store {if (!                   PVIS[S[CHOOSE].INDEX[J])//Find the factory not built {Pvis[s[choose].index[j]] = 1; Build it for (int l = head[s[choose].index[j]]; l!=-1; l = edge[l].next)//Then update the list of stores that have not been selected through the adjacency The cost of building a good factory {if (! SVIS[EDGE[L].V])//Find a store s[edge[l].v].finish + = P[S[CHOOSE].INDEX[J]].PA Y
                }}}} if (get >= tar) return get;
return 0;
    } int main () {scanf ("%d", &t);
    int tt = 1;
        while (t--) {init ();
        scanf ("%d%d%d", &n, &m, &tar);
        for (int i=1;i<=n;i++) {scanf ("%d%d", &p[i].pay, &p[i].t); } for (int i=1;i<=m;i++) {scanf ("%d%d", &s[i].pro, &s[i].
            K);
            S[i].time = S[i].cost = 0; for (int j=1;j<=s[i].
                k;j++) {scanf ("%d", &s[i].index[j]);
                Add (S[i].index[j], i);
                S[i].time = Max (S[i].time, p[s[i].index[j]].t);
            S[i].cost + = P[s[i].index[j]].pay;
        }} int l=1;
        int r=1000000000;
        int Min = INF;
        int mid;
        int ans;
            while (l<=r) {mid = (l+r)/2;
            int tmp = judge (mid);
       if (TMP)     {ans = tmp;
                min = Min (Mid, Min);
            R = mid-1;
        } else L = mid+1;
        } if (Min! = INF) printf ("Case #%d:%d%d\n", tt++, Min, ans);
    else printf ("Case #%d:impossible\n", tt++);
} return 0; }