Codeforces Round #182 (Div. 1) B. Yaroslav and Time 最短路

题目链接：

http://codeforces.com/problemset/problem/301/B

B. Yaroslav and Time

time limit per test2 seconds
memory limit per test256 megabytes
#### 问题描述
> Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
>
> A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
>
> Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
>
> Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
#### 输入
> The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
>
> The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
>
> It is guaranteed that no two stations are located at the same point.
#### 输出
> In a single line print an integer — the answer to the problem.
####样例输入
> 3 1000
> 1000
> 0 0
> 0 1
> 0 3

样例输出

2000

题意

给你二维上的n个点，两个点的时间消耗为哈密顿距离*d，且到第i个点能补充a[i]的时间，问你在起始点需要买多少的时间才能保证你能够到终点。

题解

对于每条边（u,v)，边权定为d*(u到v的哈密顿距离）-a[v]，然后由点n到点1跑spfa最短路，这里注意：由于题目给的d和a的范围，导致都是正权图。

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef __int64 LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=111;

struct Edge {
    int u,v,d;
    Edge(int u,int v,int d):u(u),v(v),d(d) {}
};

int n,D;
int arr[maxn];
int mat[maxn][maxn];
PII pt[maxn];

int dis(int i,int j) {
    return abs(pt[i].X-pt[j].X)+abs(pt[i].Y-pt[j].Y);
}

bool inq[maxn];
LL d[maxn];
LL spfa(int s) {
    queue<int> Q;
    clr(inq,0);
    for(int i=1; i<=n; i++) d[i]=INFL;
    d[s]=0,inq[s]=true,Q.push(s);
    while(!Q.empty()) {
        int u=Q.front();
        Q.pop();
        inq[u]=false;
        for(int v=1;v<=n;v++){
            if(v==u) continue;
            if(d[v]>d[u]+mat[u][v]){
                d[v]=d[u]+mat[u][v];
                if(!inq[v]){
                    Q.push(v); inq[v]=1;
                }
            }
        }
    }
    return d[1];
}

int main() {
    scf("%d%d",&n,&D);
    arr[1]=arr[n]=0;
    for(int i=2; i<n;i++) scf("%d",&arr[i]);

    for(int i=1; i<=n; i++) {
        scf("%d%d",&pt[i].X,&pt[i].Y);
    }

    for(int i=1; i<=n; i++) {
        for(int j=1; j<=n; j++) {
            if(i==j) mat[i][j]=0;
            else mat[i][j]=dis(i,j)*D-arr[j];
        }
    }

    LL ans=spfa(n);

    prf("%I64d\n",ans);

    return 0;
}

//end-----------------------------------------------------------------------