uva 11157 Dynamic Frog

题意：一只青蛙要从河的左岸到右岸找东西，找完东西要返回，河中间有呈直线的踮脚石，有两种石头，一种大石头，可以多次踩踏不沉，一种小石头，只能踩一次。

解题思路：

青蛙往返可以看作两只青蛙一起过河，求青蛙最大跨度

#include <cstdio>
#include <iostream>
#include <vector>
//#define T
using namespace std;

struct jump
{
    int cnt;
    int dis;
    jump(char ch = ' ',int diss = 0)
    {
        if(ch == 'B')
            cnt = 110;
        else
            cnt = 1;
        dis = diss;
    }
};

vector<jump> stone;

int main()
{
    #ifdef T
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
    #endif // T
    int n;
    scanf("%d",&n);
    for(int cas = 1; cas <= n; cas ++)
    {
        stone.clear();
        int a,b;
        char ch1,ch2;
        int d;
        scanf("%d %d",&a,&b);
        stone.push_back(jump('B',0));
        for(int i = 0;i<a;i++)
        {
            scanf(" %c %c %d",&ch1,&ch2,&d);   //注意前面加空格，以防读入char型换行符
           // prlongf("ch1 = %c   ch2 = %c  d = %d\n",ch1,ch2,d);
            stone.push_back(jump(ch1,d));
        }
        stone.push_back(jump('B',b));
        int ans = 0;
        int pre1 = 0;
        int pre2 = 0;
        for(int i = 0; i < stone.size(); i++)
        {
           // prlongf("stone.cnt --- %d   stone.dis ---- %d   i = %d\n",stone[i].cnt,stone[i].dis,i);
            if(stone[i].cnt > 1)
            {
                if(stone[i].dis - pre1 >ans)
                {
                    ans = stone[i].dis - pre1;

                }
                if(stone[i].dis - pre2 > ans)  // 顺序别弄错
                {
                    ans = stone[i].dis - pre2;
                }
                pre1 = stone[i].dis;
                pre2 = stone[i].dis;

            }
            else
            {
                if(pre2 > pre1)
                {
                    if(stone[i].dis - pre1 > ans)
                        ans = stone[i].dis - pre1;
                    pre1 = stone[i].dis;
                }
                else
                {
                    if(stone[i].dis - pre2 > ans)
                        ans = stone[i].dis - pre2;
                    pre2 = stone[i].dis;
                }
            }

        }
        printf("Case %d: ",cas);
        printf("%d\n",ans);
    }
   return 0;
}