条件DP UVA 672 Gangsters

题目传送门

题意：n个歹徒进饭店，可变化宽度的门，范围[0, k]，每个歹徒进门在ti时间进门，身材si，进去后有pi的成功值，问最大的成功值

分析：首先按照进门时间排序，dp[i][j] 表示第i个歹徒在门大小为j的时候进门的最大成功值，那么状态转移方程：dp[i][j] = dp[i-1][k] + a[i].p 条件是门大小从k到j的时间差小于i和ii-1人进门时间差

收获：这类有条件限制的dp转移，我分类为条件DP

代码：

/************************************************
* Author        :Running_Time
* Created Time  :2015-8-31 16:19:35
* File Name     :UVA_672.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e2 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct People   {
    int t, p, s;
    bool operator < (const People &r) const {
        return t < r.t;
    }
}a[N];
int dp[N][N];

int main(void)    {
    int T;  scanf ("%d", &T);
    while (T--) {
        int n, door, time;  scanf ("%d%d%d", &n, &door, &time);
        for (int i=1; i<=n; ++i)    scanf ("%d", &a[i].t);
        for (int i=1; i<=n; ++i)    scanf ("%d", &a[i].p);
        for (int i=1; i<=n; ++i)    scanf ("%d", &a[i].s);
        sort (a+1, a+1+n);

        memset (dp, -1, sizeof (dp));
        int ans = 0;    dp[0][0] = 0;
        for (int i=1; i<=n; ++i)    {
            for (int j=0; j<=door; ++j) {
                for (int k=0; k<=door; ++k) {
                    if (dp[i-1][k] == -1 || abs (j - k) > a[i].t - a[i-1].t)   continue;
                    if (a[i].s == j && j)   {
                        dp[i][j] = max (dp[i][j], dp[i-1][k] + a[i].p);
                    }
                    else    {
                        dp[i][j] = max (dp[i][j], dp[i-1][k]);
                    }
                    ans = max (ans, dp[i][j]);
                }
            }
        }
        printf ("%d\n", ans);
        if (T)  puts ("");
    }

    return 0;
}