ZOJ 3431 Escape!

Escape!

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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The usual stuff, after you beat the boss at the topmost floor of his castle, the evil boss triggers the auto destruct function of his castle, and you must escape! However, not so fast! You have fought really bravely and it seems you do deserve some rewards. There are N floors in the castle, and there are treasures lying about in the castle, you want to escape and get as much treasure as you can. Each piece of treasure has a value, and you would want to maximize the sum. With the auto destruction of the castle imminent, the exit stair of floor i will crumple after T_i minutes (it's still open at the T_i-th minute), you do not want to left buried in the castle, so you have to leave the i-th floor in T_i-th minute.

Input

The first line of the input is T (T≤100), the number of testcases. Then T blocks follow, each has the form: an integer N (N≤100), the number of floors. IntegersX, Y, your initial position on the N-th floor. Then N lnteger pairs, each being the X_i and Y_i, coordinate of the stairs of the i-th floor, then N lines follow each with the form: an integer M_i (0≤M_i≤5), the number of treasure of the i-th floor. Then M_i triples follow, each being a piece of treasure's X and Y coordinate and it's value. Then N integers, the i-th is T_i (T_i ≤ 1200). All the input above are given in the order of N to 1. All coordinates are in range [-1200,1200].

Output

The maximum value you can get or "I'm doomed, though I fought bravely" if you cannot escape. Assume you can only go X and Y direction and your speed is 1 unit per minute.

Sample Input

110 010 101 4 5 101000

Sample Output

10

ZOJ月赛的一个题，参考了代码 ZOJ Monthly, November 2010解题报告

题意：一个人要从房间里面逃跑，共N层，现在他在第N层，要跑到一楼出口。告诉一个在第N层的起始坐标，每一层的出口坐标。每一层都有m个宝藏，这个人在保证能逃跑出去的前提下，希望尽量多的拿到宝藏，三个数一组，分别表示宝藏的坐标和价值，每层m组。最后N个数，每一层可以待的最长时限，超过这个时间，这一层就会塌陷。注意距离为曼哈顿距离。

分析：dp题，枚举每一层对于m个宝藏拿多少，拿的顺序是什么，选择个数共有C(m,1)+C(m,2)`````+C(m,m)=2^m-1种情况，对于每种情况再枚举取的顺序。

#include <iostream>#include <stdio.h>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <cstdlib>using namespace std;const int N = 222;int n;int fk[]={1,2,4,8,16,32,64,128};int dp[N][2400];int m[N];int x[N],y[N];//每层出口坐标int xx[N][5],yy[N][5],val[N][5];//每层宝藏坐标和价值int t[N];void update(int pos,int val,int c){    for(int i=c;i<=t[pos];i++)    dp[pos][i]=max(dp[pos][i],dp[pos+1][i-c]+val);}void fun(int pos){    int a[10];    update(pos,0,abs(x[pos+1]-x[pos])+abs(y[pos+1]-y[pos]));//不拿宝藏时    int cnt;    int v;    for(int i=1;i<fk[m[pos]];i++)    {        cnt=0;v=0;        for(int j=0;j<m[pos];j++)            if(fk[j]&i)            {                v+=val[pos][j];                a[cnt++]=j;            }        int len=99999999;        int tmp;        do{            tmp=abs(xx[pos][a[0]]-x[pos+1])+abs(yy[pos][a[0]]-y[pos+1]);            for(int j=1;j<cnt;j++)            tmp+=abs(xx[pos][a[j]]-xx[pos][a[j-1]])+abs(yy[pos][a[j]]-yy[pos][a[j-1]]);            tmp+=abs(xx[pos][a[cnt-1]]-x[pos])+abs(yy[pos][a[cnt-1]]-y[pos]);            len=min(len,tmp);        }while(next_permutation(a,a+cnt));//变换到比当前字典序大一的排列，枚举取的顺序        update(pos,v,len);    }}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        for(int i=n;i>=0;i--)            scanf("%d%d",&x[i],&y[i]);        for(int i=n-1;i>=0;i--)        {            scanf("%d",&m[i]);            for(int j=0;j<m[i];j++)            scanf("%d%d%d",&xx[i][j],&yy[i][j],&val[i][j]);        }        int time=0;        for(int i=n-1;i>=0;i--)        {            scanf("%d",&t[i]);            time=max(time,t[i]);        }        for(int i=0;i<=n;i++)        for(int j=0;j<=time;j++)            dp[i][j]=-99999999;            dp[n][0]=0;        for(int i=n-1;i>=0;i--)            fun(i);        int ans=dp[0][0];        for(int i=1;i<=time;i++)        ans=max(ans,dp[0][i]);        if(ans<0)        puts("I'm doomed, though I fought bravely");        else        printf("%d\n",ans);    }

<span style="font-family: Arial, Helvetica, sans-serif;">return 0;</span>

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