題目描述

Some company is going to hold a fair in Byteland. There are n n n towns in Byteland and m m m two-way roads between towns. Of course, you can reach any town from any other town using roads.

There are k k k types of goods produced in Byteland and every town produces only one type. To hold a fair you have to bring at least s s s different types of goods. It costs d(u,v) d(u,v) d(u,v) coins to bring goods from town u u u to town v v v where d(u,v) d(u,v) d(u,v) is the length of the shortest path from u u u to v v v . Length of a path is the number of roads in this path.

The organizers will cover all travel expenses but they can choose the towns to bring goods from. Now they want to calculate minimum expenses to hold a fair in each of n n n towns.

輸入輸出格式

輸入格式：

There are 4 4 4 integers n n n , m m m , k k k , s s s in the first line of input ( 1≤n≤105 1 \le n \le 10^{5} 1≤n≤105 , 0≤m≤105 0 \le m \le 10^{5} 0≤m≤105 , 1≤s≤k≤min(n,100) 1 \le s \le k \le min(n, 100) 1≤s≤k≤min(n,100) ) — the number of towns, the number of roads, the number of different types of goods, the number of different types of goods necessary to hold a fair.

In the next line there are n n n integers a1,a2,…,an a_1, a_2, \ldots, a_n a1?,a2?,…,an? ( 1≤ai≤k 1 \le a_{i} \le k 1≤ai?≤k ), where ai a_i ai? is the type of goods produced in the i i i -th town. It is guaranteed that all integers between 1 1 1 and k k k occur at least once among integers ai a_{i} ai? .

In the next m m m lines roads are described. Each road is described by two integers u u u v v v ( 1≤u,v≤n 1 \le u, v \le n 1≤u,v≤n , u≠v u \ne v u≠v ) — the towns connected by this road. It is guaranteed that there is no more than one road between every two towns. It is guaranteed that you can go from any town to any other town via roads.

輸出格式：

Print n n n numbers, the i i i -th of them is the minimum number of coins you need to spend on travel expenses to hold a fair in town i i i . Separate numbers with spaces.

輸入輸出樣例

輸入樣例#1： 復(fù)制 5 5 4 3 1 2 4 3 2 1 2 2 3 3 4 4 1 4 5 輸出樣例#1： 復(fù)制 2 2 2 2 3 輸入樣例#2： 復(fù)制 7 6 3 2 1 2 3 3 2 2 1 1 2 2 3 3 4 2 5 5 6 6 7 輸出樣例#2： 復(fù)制 1 1 1 2 2 1 1
題意：
n 個點，m 條邊，有k種不同的物品，求從每個點出發(fā)收集s個不同物品的最短距離；
我們從每種物品 bfs；
#include<iostream> #include<cstdio> #include<algorithm> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<queue> #include<bitset> #include<ctime> #include<deque> #include<stack> #include<functional> #include<sstream> //#include<cctype> //#pragma GCC optimize(2) using namespace std; #define maxn 200005 #define inf 0x7fffffff //#define INF 1e18 #define rdint(x) scanf("%d",&x) #define rdllt(x) scanf("%lld",&x) #define rdult(x) scanf("%lu",&x) #define rdlf(x) scanf("%lf",&x) #define rdstr(x) scanf("%s",x) typedef long long ll; typedef unsigned long long ull; typedef unsigned int U; #define ms(x) memset((x),0,sizeof(x)) const long long int mod = 1e9 + 7; #define Mod 1000000000 #define sq(x) (x)*(x) #define eps 1e-4 typedef pair<int, int> pii; #define pi acos(-1.0) //const int N = 1005; #define REP(i,n) for(int i=0;i<(n);i++) typedef pair<int, int> pii; inline ll rd() {ll x = 0;char c = getchar();bool f = false;while (!isdigit(c)) {if (c == '-') f = true;c = getchar();}while (isdigit(c)) {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f ? -x : x; }ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a%b); } int sqr(int x) { return x * x; }/*ll ans; ll exgcd(ll a, ll b, ll &x, ll &y) {if (!b) {x = 1; y = 0; return a;}ans = exgcd(b, a%b, x, y);ll t = x; x = y; y = t - a / b * y;return ans; } */int n, m, k, s; vector<int>v[maxn]; vector<int>c[maxn]; int dis[100004][200]; bool vis[100004][200]; void bfs(int x) {queue<int>q;ms(vis);for (int i = 0; i < v[x].size(); i++) {q.push(v[x][i]); dis[v[x][i]][x] = 0;}while (!q.empty()) {int u = q.front(); q.pop(); vis[u][x] = 0;for (int i = 0; i < c[u].size(); i++) {int to = c[u][i];if (dis[to][x] > dis[u][x] + 1) {dis[to][x] = dis[u][x] + 1;if (!vis[to][x]) {q.push(to); vis[to][x] = 1;}}}} }int main() {//ios::sync_with_stdio(0);cin >> n >> m >> k >> s;memset(dis, 0x3f, sizeof(dis)); for (int i = 1; i <= n; i++) {int x; rdint(x);v[x].push_back(i);}for (int i = 1; i <= m; i++) {int u, V; rdint(u); rdint(V);c[V].push_back(u); c[u].push_back(V);}for (int i = 1; i <= k; i++)bfs(i);for (int i = 1; i <= n; i++) {sort(dis[i] + 1, dis[i] + 1 + k);int ans = 0;for (int j = 1; j <= s; j++)ans += dis[i][j];cout << ans << " ";}return 0; }

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轉(zhuǎn)載于:https://www.cnblogs.com/zxyqzy/p/10273636.html

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