Codeforces Round #244 (Div. 2) C. Checkposts
题目
C. Checkposts
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Your city has n junctions. There are m one-way roads between the junctions. As a mayor of the city, you have to ensure the security of all the junctions.
To ensure the security, you have to build some police checkposts. Checkposts can only be built in a junction. A checkpost at junction i can protect junction j if either i = j or the police patrol car can go to j from i and then come back to i.
Building checkposts costs some money. As some areas of the city are more expensive than others, building checkpost at some junctions might cost more money than other junctions.
You have to determine the minimum possible money needed to ensure the security of all the junctions. Also you have to find the number of ways to ensure the security in minimum price and in addition in minimum number of checkposts. Two ways are different if any of the junctions contains a checkpost in one of them and do not contain in the other.
Input
In the first line, you will be given an integer n, number of junctions (1 ≤ n ≤ 105). In the next line, n space-separated integers will be given. The ith integer is the cost of building checkpost at the ith junction (costs will be non-negative and will not exceed 109).
The next line will contain an integer m (0 ≤ m ≤ 3·105). And each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n; u ≠ v). A pair ui, vi means, that there is a one-way road which goes from ui to vi. There will not be more than one road between two nodes in the same direction.
Output
Print two integers separated by spaces. The first one is the minimum possible money needed to ensure the security of all the junctions. And the second one is the number of ways you can ensure the security modulo 1000000007 (109 + 7).
Sample test(s)
Input
3 1 2 3 3 1 2 2 3 3 2
Output
3 1
Input
5 2 8 0 6 0 6 1 4 1 3 2 4 3 4 4 5 5 1
Output
8 2
Input
10 1 3 2 2 1 3 1 4 10 10 12 1 2 2 3 3 1 3 4 4 5 5 6 5 7 6 4 7 3 8 9 9 10 10 9
Output
15 6
Input
2 7 91 2 1 2 2 1
Output
7 1
解析
图的强连通分支查找，可以使用双向遍历的方法求解，也可使用tarjan方法求解，这里选择tarjan
答案
#include <cstring>
#include <iostream>
#include <vector>
#include <stack>
using namespace std;
#define CLR(vertex, value) memset(vertex, value, sizeof(vertex))
/*used for solve*/
#define MAX_CHECKPOST 101000
#define MAX_WAY 301000
#define MAX_COST ((int)( (~(unsigned int)0) >> 1))
#define MOD 1000000007
vector<int> graph[MAX_CHECKPOST]; /*record the graph*/
int cost_table[MAX_CHECKPOST]; /*checkpost buildcost*/
int tot_checkpost; /*tot checkpost*/
int tot_way; /*tot way*/
/*store info for strong connected*/
//int strongconnected_table[MAX_CHECKPOST]; /*strongconnected_table id*/
int min_value[MAX_CHECKPOST]; /*min cost in this strong connected*/
int min_cnt[MAX_CHECKPOST]; /*tot pos in min value in this strongconnected*/
/*used for tarjan*/
int dfs_clock, low[MAX_CHECKPOST], dfn[MAX_CHECKPOST], strongconnected_group;
bool in_stack[MAX_CHECKPOST];
stack<int> pos_stack;
static void tarjan(int curr_pos);
static void init_tarjan();
static void run_tarjan();
static void read_input();
static void build_output();
/**
* @brief tarjan
* @param[in] curr_pos in graph
*
*/
void tarjan(int curr_pos){
int next_pos;
dfn[curr_pos] = low[curr_pos] = dfs_clock++;
pos_stack.push(curr_pos);
in_stack[curr_pos] = true;
for(int way_id = 0; way_id < graph[curr_pos].size(); way_id++){
next_pos = graph[curr_pos][way_id];
if(!dfn[next_pos]){ /*first visit*/
tarjan(next_pos);
low[curr_pos] = min(low[curr_pos], low[next_pos]);
}else if(in_stack[next_pos]){ /*is in stack*/
low[curr_pos] = min(low[curr_pos], dfn[next_pos]);
}
}
if(low[curr_pos] == dfn[curr_pos]){
int strongconnected_root = curr_pos;
int strongconnected_child;
strongconnected_group++;
do{
strongconnected_child = pos_stack.top();
//strongconnected_table[strongconnected_child] = strongconnected_group;
pos_stack.pop();
in_stack[strongconnected_child] = false;
if(cost_table[strongconnected_child] < min_value[strongconnected_group]){
min_value[strongconnected_group] = cost_table[strongconnected_child];
min_cnt[strongconnected_group] = 1;
}else if( cost_table[strongconnected_child] == min_value[strongconnected_group]){
min_cnt[strongconnected_group]++;
}
#if 0
cout << " strongconnected_group = " << strongconnected_group << endl;
cout << " strongconnected_child = " << strongconnected_child << endl;
cout << " min_value = " << min_value[strongconnected_group] << endl;
cout << " min_cnt = " << min_cnt[strongconnected_group] << endl;
cout << endl;
#endif
}while( strongconnected_child != strongconnected_root);
}
}
/**
* @brief init for tarjan
*
*/
void init_tarjan(){
dfs_clock = 1;
strongconnected_group = 0;
for(int curr_checkpost = 1; curr_checkpost <= tot_checkpost; curr_checkpost++){
min_value[curr_checkpost] = MAX_COST;
}
CLR(in_stack, false);
}
/**
* @brief run tarjan
* @param[in] curr_pos in graph
*
*/
void run_tarjan(){
int curr_checkpost;
init_tarjan();
for(curr_checkpost = 1; curr_checkpost <= tot_checkpost; curr_checkpost++){
if(!dfn[curr_checkpost]){
tarjan(curr_checkpost);
}
}
}
/**
* @brief build output by tarjan
*
*/
void build_output()
{
long long int totvalue = 0;
long long int totpos = 1;
int sc_id;
for(sc_id = 1; sc_id <= strongconnected_group; sc_id++){
#if 0
cout << "sc_id = " << sc_id << endl;
cout << "min_cnt = " << min_cnt[sc_id] << endl;
#endif
totvalue += min_value[sc_id];
totpos *= min_cnt[sc_id];
if(totpos > MOD){
totpos %= MOD;
}
}
cout << totvalue << " " << totpos << endl;
}
/**
* @brief read input for problem
*
*/
void read_input()
{
int curr_checkpost, curr_way, from_pos, to_pos;
cin >> tot_checkpost;
for(curr_checkpost = 1; curr_checkpost <= tot_checkpost; curr_checkpost++){ /*record all cost*/
cin >> cost_table[curr_checkpost];
}
cin >> tot_way;
for(curr_way = 1; curr_way <= tot_way; curr_way++){ /*build graph*/
cin >> from_pos;
cin >> to_pos;
graph[from_pos].push_back(to_pos);
}
}
int main()
{
read_input();
run_tarjan();
build_output();
return 0;
}