Conditional DP UVA 672 gangsters

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Test instructions: N gangsters into the hotel, can vary the width of the door, the range [0, K], each gangster entered the door in ti time, stature si, after entering the success value of Pi, ask the maximum success value

Analysis: First according to the order of the door time, Dp[i][j] indicates the first gangster in the door size of J when the maximum success of the door, then the state transfer equation: dp[i][j] = dp[i-1][k] + A[I].P condition is the gate size from K to J time difference is less than I and ii-1 people door difference

Harvesting: This type of conditional DP transfer I classify as conditional DP

Code:

/************************************************* author:running_time* Created time:2015-8-31 16:19:35* File Na Me: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 | 1typedef 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", &AMP;A[I].P);        for (int i=1; i<=n; ++i) scanf ("%d", &AMP;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;}

　　

Conditional DP UVA 672 gangsters