AcWing 324. 贿赂FIPA

标签：二维   ace   problem   als   next   ipa   string   line   计划   

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;
    }
}

AcWing 324. 贿赂FIPA

标签：二维   ace   problem   als   next   ipa   string   line   计划   

原文地址：https://www.cnblogs.com/dmoransky/p/12380439.html