AcWing 324. 贿赂FIPA

题目链接

大型补档计划

\(f[i][j]\) 表示第 \(i\) 个国家,获得 \(j\) 个国家支持,用的最少花费

\(f[i][0] = 0\)
\(f[i][sz[i]] = w[i]\)

对于每条边 \((u, v)\)
枚举 \(u\) 的第二维 \(j\)\(v\) 的第二维 \(k\) \((k <= j)\)
\(f[u][j] = min(f[u][j], f[v][k] + f[u][j - k])\)

#include <cstdio>
#include <iostream>
#include <map>
#include <string>
#include <cstring>
#include <sstream>
using namespace std;
const int N = 205;
int n, m;
int head[N], numE = 0, out[N], idx = 1;
map<string, int> st;
int sz[N], f[N][N], w[N];
struct E{
	int next, v;
} e[N];
void add(int u, int v) {
	e[++numE] = (E) { head[u], v };
	head[u] = numE;
}

void dfs(int u) {
    sz[u] = 1;
	f[u][0] = 0;
	for (int i = head[u]; i; i = e[i].next) {
		int v = e[i].v;
		dfs(v);
		sz[u] += sz[v];
		for (int j = m; j >= 1; j--)
			for (int k = 1; k <= j; k++)
				f[u][j] = min(f[u][j], f[u][j - k] + f[v][k]);
	}
	f[u][min(sz[u], m)] = min(f[u][min(sz[u], m)], w[u]); 
	for (int i = m - 1; i >= 1; i--) f[u][i] = min(f[u][i], f[u][i + 1]);
}

int main() {
	ios::sync_with_stdio(false);
	while (cin >> n >> m) {
		numE = 0; idx = 1;
		memset(head, 0, sizeof head);
		memset(f, 0x3f, sizeof f);
		memset(out, 0, sizeof out);
		st.clear();
		for (int i = 1; i <= n; i++) {
			int val; string s, b, c; 
			cin >> c >> val;
			if (!st[c]) st[c] = ++idx;
			w[st[c]] = val;
			int a = st[c];
			getline(cin, s);
			stringstream ss(s);
			while (ss >> b) {
				if (!st[b]) st[b] = ++idx;
				add(a, st[b]); out[st[b]]++; 
			}
		}
 		for (int i = 2; i <= idx; i++) if (!out[i]) add(1, i);
 		w[1] = 1e9; dfs(1);
 		cout << f[1][m] << endl;
	}
}
posted @ 2020-02-28 21:44  DMoRanSky  阅读(142)  评论(0编辑  收藏  举报