POJ 1661 Help Jimmy

题解：jimmy跳到下一个平台，总是从当前平台的最左侧或者最右侧跳下去的。dp[i][0]表示从i个平台左边跳下去，dp[i][1]表示从第i平台右边跳下去。

#include <iostream>#include <string.h>#include <cmath>#include <cstdio>#include <algorithm>using namespace std;const int maxn = 1005;#define inf 0x3f3f3f3fint t,n,x,y,maxh;int dp[maxn][2];struct Node{    int x1,x2,h;}e[maxn];bool cmp(Node t1,Node t2){    return t1.h<t2.h;}void lefttime(int i){           //从平台左边跳下去的时间    int k = i-1;    while(k>0&&e[i].h-e[k].h<=maxh){        if(e[k].x1<=e[i].x1&&e[k].x2>=e[i].x1){            dp[i][0] = min(dp[k][0]+e[i].x1-e[k].x1,dp[k][1]+e[k].x2-e[i].x1)+e[i].h-e[k].h;            return;        }        k--;    }    if(e[i].h<=maxh)        dp[i][0] = e[i].h;    else        dp[i][0] = inf;}void righttime(int i){          //从平台右边跳下去的时间    int k = i-1;    while(k>0&&e[i].h-e[k].h<=maxh){        if(e[k].x1<=e[i].x2&&e[k].x2>=e[i].x2){            dp[i][1] = min(dp[k][0]+e[i].x2-e[k].x1,dp[k][1]+e[k].x2-e[i].x2)+e[i].h-e[k].h;            return;        }        k--;    }    if(e[i].h<=maxh)        dp[i][1] = e[i].h;    else        dp[i][1] = inf;}int solve(){    for(int i=1;i<=n+1;i++){        lefttime(i);        righttime(i);    }    return min(dp[n+1][0],dp[n+1][1]);}int main(){    scanf("%d",&t);    while(t--){        scanf("%d%d%d%d",&n,&x,&y,&maxh);        for(int i=1;i<=n;i++){            scanf("%d%d%d",&e[i].x1,&e[i].x2,&e[i].h);        }        e[n+1].x1 = e[n+1].x2 = x;        e[n+1].h = y;        e[0].x1=-2000;        e[0].x2=2000;        e[0].h=0;        sort(e,e+n+1,cmp);        printf("%d\n",solve());    }    return 0;}