Simple dfs hdu 4536 XCOM Enemy Unknown

Question meaning:

In n countries, each country has a fear value, and each country belongs to one continent. Aliens Attack k times. Each time they attack three countries in different continents, they will ask you the maximum number of times you can defend against attacks as a consortium.

Each attack can support a country. After the attack, the fear Value of the country is-2, but the fear value of the other two countries is + 2, in addition, the fear value of all other countries in the continent where the other two countries are located is + 1.

When the fear value of a country is greater than 5, the task fails and cannot continue to resist the attack. Each country's fear value is minimized to 1.

Solution:

Dfs brute force search.

Three options are available each time. Select the country to be supported.

Code:

 

# Include <iostream> # include <cmath> # include <cstdio> # include <cstdlib> # include <string> # include <cstring> # include <algorithm> # include <vector> # include <map> # include <set> # include <stack> # include <list> # include <queue> # define eps 1e-6 # define INF 0x1f1f1f1f # define PI acos (-1.0) # define ll _ int64 # define lson l, m, (rt <1) # define rson m + 1, r, (rt <1) | 1 // # pragma comment (linker, "/STACK: 1024000000,1024000000") using namespace std;/* freopen ("data. in "," r ", stdin); freopen (" data. out "," w ", stdout); */int n, m, k; int st [20], fear [20]; vector <int> myv [6]; struct Node {int a [3];} fi [110]; bool flag; int ans; void dfs (int cur, int * afr) {if (flag) return; for (int I = 0; I <n; I ++) {if (afr [I]> 5) // more than 5, it indicates that the last time has failed, therefore, it is cur-2 {ans = max (cur-2, ans); return ;}} if (cur> k) // can resist all attacks {flag = true; ans = k; return;} int tmp [20]; for (int I = 0; I <3; I ++) // select the protected country {// memcpy (tmp, afr, sizeof (afr); for (int j = 0; j <n; j ++) tmp [j] = afr [j]; tmp [fi [cur]. a [I]-= 2; // reduce the national fear value by 2 if (tmp [fi [cur]. a [I] <1) tmp [fi [cur]. a [I] = 1; for (int j = 1; j <= 2; j ++) // process the other two countries {int k = (I + j) % 3; int sta = st [fi [cur]. a [k]; for (int p = 0; p <myv [sta]. size (); p ++) {tmp [myv [sta] [p] + = 1;} tmp [fi [cur]. a [k] ++;} // printf ("cur: % d I: % d \ n", cur, I);/* for (int j = 0; j <n; j ++) printf ("% d", tmp [j]); putchar ('\ n'); */dfs (cur + 1, tmp) ;}} int main () {// printf ("% lf \ n", pow (5.0, 16.0); int t; scanf ("% d ", & t); for (int ca = 1; ca <= t; ca ++) {scanf ("% d", & n, & m, & k); for (int I = 0; I <m; I ++) myv [I]. clear (); for (int I = 0; I <n; I ++) {scanf ("% d", & st [I]); // The continent myv [st [I]. push_back (I); // all countries in the continent} for (int I = 0; I <n; I ++) scanf ("% d ", & fear [I]); // The fear value for (int I = 1; I <= k; I ++) for (int j = 0; j <3; j ++) scanf ("% d", & fi [I]. a [j]); // alien attack country ans = 0; flag = false; dfs (1, fear); printf ("Case # % d: % d \ n ", ca, ans);} return 0 ;}