Hdu4939 stupid tower defense (DP)

2014 + school 7 second-level questions

4939

Stupid tower defenseTime Limit: 12000/6000 MS (Java/others) memory limit: 131072/131072 K (Java/others) total submission (s): 366 accepted submission (s): 88Problem description FSF is addicted to a stupid tower defense game. the goal of tower Defense Games is to try to stop enemies from crossing a map by building traps to slow them down and towers which shoot at them as they pass. The map is a line, which has n unit length. we can build only one tower on each unit length. the enemy takes t seconds on each unit length. and there are 3 kinds of tower in this game: The Red Tower, the Green Tower and the blue tower. The Red Tower damage on the enemy X points per second when he passes through the tower. The Green Tower damage on the enemy y points per second after he passes through the tower. The blue tower let the enemy go slower than before (that is, the enemy takes more Z second to pass an unit length, also, after he passes through the tower .) of course, if you are already pass through M green towers, you shoshould have got M * y damage per second. the same, if you are already pass through K blue towers, the enemy shoshould have took T + K * z seconds every unit length. FSF now wants to know the maximum damage the enemy can get. Input there are multiply test cases. The first line contains an integer T (t <= 100), indicates the number of cases. Each test only contain 5 integers n, x, y, z, T (2 <= n <= 1500,0 <= x, y, z <= 60000,1 <= T <= 3) output for each case, You shoshould output "case # C:" First, where C indicates the case number and counts from 1. then output the answer. for each test only one line which have one integer, the answer to this question. sample input12 4 3 2 1 sample outputcase #1: 12HintFor the first sample, the first tower is blue tower, and the second is Red Tower. so, the total damage is 4*(1 + 2) = 12 damage points. authoruestc source 2014 multi-university training contest 7 recommendwe have carefully selected several similar problems for you: 4944 4943 4942 4941 4940

Question:Tower guard, it's strange that the initial time for each block is T. A tower can be built for each piece. There are three types of towers. One is the red tower, and the other is X output per second when it passes through the current grid. The other is a green Tower, which is attacked by virus Y every second after it passes through the current grid; one is the blue tower. The time that passes through the current grid increases by Z seconds. The Green Tower and the blue tower can be combined to find the highest output.

Problem: DP.

Thinking questions, you can think of the red tower to put the last is the optimal solution, you can prove: if there is a red tower behind the red tower, exchange them will certainly have a better solution.

In this way, we will enumerate the number of non-red towers and the number of green towers. f [I] [J] indicates that there are I non-red towers in front of them, and when there are j green towers, the blue-green tower outputs and in the whole road section.

Because I and j are known, the output of the red tower behind them is also fixed (only as I and j change, and the output of the Red Tower remains unchanged when IJ Is fixed ), therefore, we use DP to obtain the largest f [I] [J] of various IJ, and ANS update each time.

Because the number seems to be a little large, it is better to use a queue that will be rolled out with a super commit.

1 // # pragma comment (linker, "/Stack: 102400000,102400000 ") 2 # include <cstdio> 3 # include <cmath> 4 # include <iostream> 5 # include <cstring> 6 # include <algorithm> 7 # include <cmath> 8 # include <map> 9 # include <set> 10 # include <stack> 11 # include <queue> 12 using namespace STD; 13 # define ll long long14 # define usll unsigned ll15 # define MZ (array) memset (array, 0, sizeof (array) 16 # define minf (array) memset (ARR Ay, 0x3f, sizeof (array) 17 # define rep (I, n) for (I = 0; I <(n); I ++) 18 # define for (I, X, n) for (I = (x); I <= (n); I ++) 19 # define RD (X) scanf ("% d", & X) 20 # define RD2 (x, y) scanf ("% d", & X, & Y) 21 # define rd3 (x, y, z) scanf ("% d", & X, & Y, & Z) 22 # define Wn (X) prllf ("% d \ n", x); 23 # define re freopen ("D. in "," r ", stdin) 24 # define we freopen (" 1biao. out "," W ", stdout) 25 ll max (ll x, ll y) {26 return x> Y? X: Y; 27} 28 ll f [2] [1555] [1555]; // green blue29 30 int main () {31 ll t, CAS = 1; 32 ll N, T; 33 ll x, y, z; 34 ll now, pre; 35 ll I, j, k; 36 scanf ("% i64d", & T ); 37 while (t --) {38 scanf ("% i64d % i64d % i64d % i64d", & N, & X, & Y, & Z, & T ); 39 memset (F, 0, sizeof (f); 40 now = 0; 41 pre = 1; 42 ll ans = x * T * n; 43 for (I = 1; I <= N; I ++) {// 1st to the 44 for (j = 0; j <= I; j ++) {// 1st to I have J green45 K = I-j; // 1st to I have k blue46 if (k> 0) {// 47 F [now] [J] [k] = f [pre] [J] [k-1] + J * y * z *( n-I ); /// put this blue. The next n-I lattice increases the Z time by 48} 49. If (j> 0) {// Green 50 F [now] [J] [k] = max (F [now] [J] [K], f [pre] [J-1] [k] + y * (t + K * z) * (N-I )); /// add y damage every second when you put this green color. 51} 52 ans = max (ANS, F [now] [J] [k] + x * (N-I) * (t + K * z); 53} 54 pre = now; 55 now ^ = 1; 56} 57 printf ("case # % i64d: % i64d \ n ", CAS ++, ANS); 58} 59 return 0; 60}

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