【AMPPZ 2014】 The Captain

【题目链接】

点击打开链接

【算法】

按x轴排序，将相邻点连边

按y轴排序，将相邻点连边

然后对这个图跑最短路就可以了，笔者用的是dijkstra算法

【代码】

#include<bits/stdc++.h>
using namespace std;
#define MAXN 200000

struct info {
        int id,x,y;
} a[MAXN+];

int N,i;
int dist[MAXN+];
vector< pair<int,int> > E[MAXN+];

template <typename T> inline void read(T &x) {
        int f = ; x = ;
        char c = getchar();
        for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
        for (; isdigit(c);  c = getchar()) x = (x << ) + (x << ) + c - '';
        x *= f;
}
template <typename T> inline void write(T x) {
        if (x < ) { putchar('-'); x = -x; }
        if (x > ) write(x/);
        putchar(x%+'');
}
template <typename T> inline void writeln(T x) {
        write(x);
        puts("");
}

inline bool cmp1(info a,info b) { return a.x < b.x; }
inline bool cmp2(info a,info b) { return a.y < b.y; }

inline void add_edge(int x,int y,int d) {
        E[x].push_back(make_pair(y,d));
        E[y].push_back(make_pair(x,d));
}

inline void dijkstra() {
        int i,x,to,cost;
        static int vis[MAXN+];
        priority_queue< pair<int,int> > q;
        for (i = ; i <= N; i++) dist[i] = INT_MAX;
        q.push(make_pair(,));
        while (!q.empty()) {
                x = q.top().second; q.pop();
                if (vis[x]) continue;
                for (i = ; i < E[x].size(); i++) {
                        to = E[x][i].first;
                        cost = E[x][i].second;
                        if (dist[x] + cost < dist[to]) {
                                dist[to] = dist[x] + cost;
                                q.push(make_pair(-dist[to],to));
                        }
                }
        }
}

int main() {

        read(N);
        for (i = ; i <= N; i++) {
                read(a[i].x); read(a[i].y);
                a[i].id = i;
        }
        sort(a+,a+N+,cmp1);
        for (i = ; i <= N; i++) add_edge(a[i-].id,a[i].id,a[i].x-a[i-].x);
        sort(a+,a+N+,cmp2);
        for (i = ; i <= N; i++) add_edge(a[i-].id,a[i].id,a[i].y-a[i-].y);

        dijkstra();
        writeln(dist[N]);

        return ;

}