hdu1690Bus System--解題報告

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題意：有一個公交系統的收費標準如下表：

然後問：給出 這些L1~4 & C1~4的值，然後 N個站，列出每個站的X座標，然後詢問M次，問兩個月台的最小花費題解：那麼這裡很明顯是最短路問題，有一點的麻煩就在於建圖，那麼我們可以對於所有的點，用兩個for迴圈，算出兩兩之間的距離，就可以得到花費是多少，同時建邊，然後對於每次詢問的點，我們就spfa一次就OK

<span style="font-size:14px;">#include <iostream>#include <cstdio>#include <cmath>#include <queue>#include <cstring>using namespace std;#define INF 0xffffffffffffff#define MAX 105#define LL __int64int N,M;LL L1,L2,L3,L4,C1,C2,C3,C4;LL X[MAX];struct Edge{    int to,next;    LL cost;}edge[MAX*MAX];int head[MAX],tol;void add(int u,int v,LL cost){    edge[tol].to = v;    edge[tol].cost = cost;    edge[tol].next = head[u];    head[u] = tol++;}void del() //處理建邊{    LL cost,dis;    for(int i = 1; i <= N; i ++){        for(int j = i+1; j <= N; j ++){            if(X[i] > X[j]) dis = X[i]-X[j];            else dis = X[j]-X[i];            if(dis > L4) cost = INF;            else if(dis > L3) cost = C4;            else if(dis > L2) cost = C3;            else if(dis > L1) cost = C2;            else cost = C1;            add(i,j,cost);            add(j,i,cost);        }    }}LL dis[MAX];bool flag[MAX];LL spfa(int src,int D){    for(int i = 1; i <= N; i ++) dis[i] = INF;    memset(flag,false,sizeof(flag));    dis[src] = 0;    flag[src] = true;    queue<int>q;    q.push(src);    while(!q.empty())    {        int u = q.front(); q.pop();        flag[u] = false;        for(int i = head[u]; i != -1; i = edge[i].next)        {            int v = edge[i].to; LL cost = edge[i].cost;            if(cost + dis[u] < dis[v])            {                dis[v] = cost+dis[u];                if(!flag[v])                {                    q.push(v);                    flag[v] = true;                }            }        }    }    return dis[D];}int main(){    int T;    scanf("%d",&T);    for(int cas = 1; cas <= T; cas ++)    {        scanf("%I64d%I64d%I64d%I64d%I64d%I64d%I64d%I64d",&L1,&L2,&L3,&L4,&C1,&C2,&C3,&C4);        scanf("%d%d",&N,&M);        for(int i = 1; i <= N; i ++) scanf("%I64d",&X[i]);        memset(head,-1,sizeof(head));        tol = 0;        del();        printf("Case %d:\n",cas);        int a,b;        LL ans = 0;        for(int i = 0; i < M; i ++)        {            scanf("%d%d",&a,&b);            ans = spfa(a,b);            if(ans >= INF)               printf("Station %d and station %d are not attainable.\n",a,b);            else               printf("The minimum cost between station %d and station %d is %I64d.\n",a,b,ans);        }    }    return 0;}</span>

那麼這裡的話，還要注意的是 因為座標值比較大，我們用 64位來儲存