Acwing 903 昂贵的聘礼

题目描述

思路

首先从题目中抽象出来模型，即求以编号\(1\)的点为终点，求最短距离。终点确定了，但我们并没有确定起点，可以使用一个虚拟源点作为起点，该虚拟源点到其他点的距离为够买该编号商品原本的价钱。这样问题就变成了以虚拟源点\(S\)为起点，以编号\(1\)为终点的最短路。由于等级限制只有\(100\)，所以我们可以枚举等级区间，对于每个等级区间的最短路取\(min\)即可。由于\(1\)号商品是一定要买的，所以我们枚举的区间为\([level_{1}-m,level_{1}]\)。

AC代码

#include "iostream"
#include "cstring"
#include "string"
#include "vector"
#include "cmath"
#include "algorithm"
#include "map"
#include "set"
#include "queue"
#include "stack"
#include "cassert"
#include "unordered_map"
#include "sstream"
#include "cstdio"

using namespace std;

#define fi first
#define se second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) a.begin(),a.end()
#define rep(x,l,u) for(ll x=l;x=u;x--)
#define sz(x) x.size()
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define seteps(N) setprecision(N)
#define uni(x) sort(all(x)), x.erase(unique(all(x)), x.end())
#define lson (ind<<1)
#define rson (ind<<1|1) 

typedef long long ll;
typedef unsigned long long ull;
typedef __int128 lll;
typedef pair PII;
typedef pair PCC;
typedef pair PDD;
typedef pair PLL;
typedef pair PIII;


struct Scanner {
 
    bool hasNext = 1;
    bool hasRead = 1;
 
    int nextInt() {
        hasRead = 0;
        int res = 0;
        char flag = 1, ch = getchar();
        while(ch != EOF && !isdigit(ch)) {
            hasRead = 1;
            flag = (ch == '-') ? -flag : flag;
            ch = getchar();
        }
        while(ch != EOF && isdigit(ch)) {
            hasRead = 1;
            res = res * 10 + (ch - '0');
            ch = getchar();
        }
        if(ch == EOF)
            hasNext = 0;
        return res * flag;
    }
 
    ll nextLL() {
        hasRead = 0;
        ll res = 0;
        char flag = 1, ch = getchar();
        while(ch != EOF && !isdigit(ch)) {
            hasRead = 1;
            flag = (ch == '-') ? -flag : flag;
            ch = getchar();
        }
        while(ch != EOF && isdigit(ch)) {
            hasRead = 1;
            res = res * 10 + (ch - '0');
            ch = getchar();
        }
        if(ch == EOF)
            hasNext = 0;
        return res * flag;
    }
 
    char nextChar() {
        hasRead = 0;
        char ch = getchar();
        while(ch != EOF && isspace(ch)) {
            hasRead = 1;
            ch = getchar();
        }
        if(ch == EOF)
            hasNext = 0;
        return ch;
    }
 
    int nextString(char *str) {
        hasRead = 0;
        int len = 0;
        char ch = getchar();
        while(ch != EOF && isspace(ch)) {
            hasRead = 1;
            ch = getchar();
        }
        while(ch != EOF && !isspace(ch)) {
            hasRead = 1;
            str[++len] = ch;
            ch = getchar();
        }
        str[len + 1] = 0;
        if(ch == EOF)
            hasNext = 0;
        return len;
    }
 
} sc;
 
ll rd() {
    ll x = sc.nextLL();
    return x;
}
 
void rd(int &x) {
    x = sc.nextInt();
}
 
void rd(ll &x) {
    x = sc.nextLL();
}
 
void rd(char &x) {
    x = sc.nextChar();
}
 
void rd(char* x) {
    sc.nextString(x);
}
 
template
void rd(pair &x) {
    rd(x.first);
    rd(x.second);
}
 
template
void rd(T *x, int n) {
    for(int i = 1; i <= n; ++i)
        rd(x[i]);
}

template
void rd(vector &x,int n){
    for(int i = 1; i <= n; ++i)
        rd(x[i]);
}
 
void printInt(int x) {
    if(x < 0) {
        putchar('-');
        x = -x;
    }
    if(x >= 10)
        printInt(x / 10);
    putchar('0' + x % 10);
}
 
void printLL(ll x) {
    if(x < 0) {
        putchar('-');
        x = -x;
    }
    if(x >= 10)
        printLL(x / 10);
    putchar('0' + x % 10);
}
 
void pr(int x, char ch = '\n') {
    printInt(x);
    putchar(ch);
}
 
void pr(ll x, char ch = '\n') {
    printLL(x);
    putchar(ch);
}

//#define LOCAL
template
void pr(pair x, char ch = '\n') {
#ifdef LOCAL
    putchar('<');   
    pr(x.first, ' ');
    pr(x.second, '>');
    putchar(ch);
    return;
#endif //LOCAL
    pr(x.first, ' ');
    pr(x.second, ch);
}
template
void pr(T *x, int n) {
    for(int i = 1; i <= n; ++i)
        pr(x[i], " \n"[i == n]);
}
 
template
void pr(vector &x) {
    int n = x.size();
    for(int i = 1; i <= n - 1; ++i)
        pr(x[i], " \n"[i == n - 1]);
}

const int N=105;
const int M=205;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const int mod1=998244353;
const lll oone=1;
const double eps=1e-6;
const double pi=acos(-1);

int n,m;
int g[N][N],level[N],d[N];
bool st[N];
int dijkstra(int down,int up){
    mst(d,INF);
    mst(st,false);
    d[0]=0;
    rep(i,0,n){
        int t=-1;
        rep(j,0,n+1) if(!st[j] && (t==-1 || d[t]>d[j])) t=j;
        rep(j,1,n+1){
            if(level[j]>=down && level[j]<=up){
                d[j]=min(d[j],d[t]+g[t][j]);
            }
        }
        st[t]=true;
    }

    return d[1];
}
void solve(){
    rd(m);
    rd(n);
    mst(g,INF);
    rep(i,1,n+1) g[i][i]=0;
    rep(i,1,n+1){
        int p,l,x;
        rd(p);rd(l);rd(x);
        g[0][i]=min(p,g[0][i]);
        level[i]=l;
        rep(j,0,x){
            int idx,price;
            rd(idx);rd(price);
            g[idx][i]=min(g[idx][i],price);
        }
    }
    int ans=INF;
    rep(i,level[1]-m,level[1]+1){
        ans=min(ans,dijkstra(i,i+m));
    }
    pr(ans);
}
int main(){
    //IOS;
    //freopen("data.in", "r", stdin);
    //freopen("data.out", "w", stdout);
    //int t;rd(t);
    //rep(i,0,t){
        //printf("Case #%d: ", i+1);
        solve();
    //}
    return 0;
}