POJ - 1062 - 昂贵的聘礼 = Dijkstra

http://poj.org/problem?id=1062

题意：有100个物品，每个物品有一个价格值，一个地位值，和他可以用别的物品来补差价换。求换到1号物品的最小代价。

意思就是给一个图，支付起点的点权，然后走边权走到1号点，求最小的代价，其中路上经过的地位值的差不能超过题目的限制。

最暴力的做法，枚举地位的差，每次在反图上跑一次Dijkstra就好。这样貌似进行了多余的拷贝。

#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<map>
#include<set>
#include<stack>
#include<string>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;

vector<pair<int, int> >G0[105];
vector<pair<int, int> >G[105];

struct Item {
    int id;
    int st;
    int val;
} item[105];

int dis[105];
bool vis[105];

priority_queue<pair<int, int> > pq;
void dijkstra(int s, int n) {
    while(!pq.empty())
        pq.pop();
    for(int i = 1; i <= n; ++i) {
        dis[i] = 0x3f3f3f3f;
        vis[i] = 0;
    }
    dis[s] = 0;
    pq.push({-dis[s], s});
    while(!pq.empty()) {
        int u = pq.top().second;
        pq.pop();
        if(vis[u])
            continue;
        vis[u] = 1;
        for(int i = 0; i < G[u].size(); ++i) {
            int v = G[u][i].first, w = G[u][i].second;
            if(dis[v] > dis[u] + w) {
                dis[v] = dis[u] + w;
                pq.push({-dis[v], v});
            }
        }
    }
}

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    int m, n;
    scanf("%d%d", &m, &n);
    for(int i = 1; i <= n; ++i) {
        item[i].id = i;
        scanf("%d%d", &item[i].val, &item[i].st);
        int x;
        scanf("%d", &x);
        while(x--) {
            int v, w;
            scanf("%d%d", &v, &w);
            G0[i].push_back({v, w});
        }
    }
    int minval = 0x3f3f3f3f;
    for(int cst = item[1].st - m; cst <= item[1].st; ++cst) {
        for(int i = 1; i <= n; ++i)
            G[i].clear();
        int dst = cst + m;
        for(int i = 1; i <= n; ++i) {
            if(item[i].st >= cst && item[i].st <= dst) {
                G[i] = G0[i];
            }
        }

        dijkstra(1, n);
        for(int i = 1; i <= n; ++i) {
            dis[i] += item[i].val;
            if(item[i].st >= cst && item[i].st <= dst)
                minval = min(minval, dis[i]);
        }
    }
    printf("%d\n", minval);
}

标签：pq,int,1062,105,item,Dijkstra,POJ,include,dis
来源： https://www.cnblogs.com/Inko/p/11715263.html