任务
看了很久并没有懂，以下是标程的分析。
好难啊啊啊啊啊
1 #include <iostream> 2 #include <algorithm> 3 #include <cstring> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <stack> 8 using namespace std; 9 10 const int maxn = 3005; 11 int a[maxn],b[maxn],dp[2][maxn*2]; 12 #define inf 0x3f3f3f3f 13 int main() { 14 int t; 15 scanf("%d",&t); 16 while(t--) { 17 int n; 18 scanf("%d",&n); 19 for(int i = 1 ; i <= n ; i ++) scanf("%d %d",&a[i],&b[i]); 20 int off = 3000; 21 memset(dp,inf,sizeof dp); 22 dp[0][3000] = 0; 23 int k = 0; 24 for(int i = 1; i <= n ; i ++) { 25 for(int j = 0 ; j < 6001 ; j ++) { 26 if(dp[k][j] != inf) { 27 if(j > off) { 28 dp[1-k][off+a[i]] = min(dp[k][j]+a[i],dp[1-k][off+a[i]]); 29 dp[1-k][j-b[i]] = min( dp[1-k][j-b[i]], dp[k][j] + max(0,off-j+b[i]) ); 30 } 31 else { 32 dp[1-k][j +a[i] ] = min(dp[1-k][j+a[i]] ,dp[k][j] + max(0,j+a[i]-off)); 33 dp[1-k][off-b[i]] = min(dp[1-k][off-b[i]] ,dp[k][j]+b[i]); 34 35 } 36 } 37 } 38 memset(dp[k],inf,sizeof dp[k]); 39 k = 1 - k; 40 } 41 int ans = inf ; 42 for(int i = 0 ; i <= 6000; i++) { 43 if(dp[k][i] != inf) { 44 ans = min(ans,dp[k][i] ); 45 } 46 } 47 printf("%d\n",ans); 48 } 49 return 0; 50 }
大佬的：
1 // 2 // Created by TaoSama on 2016-05-17 3 // Copyright (c) 2016 TaoSama. All rights reserved. 4 // 5 #pragma comment(linker, "/STACK:102400000,102400000") 6 #include <algorithm> 7 #include <cctype> 8 #include <cmath> 9 #include <cstdio> 10 #include <cstdlib> 11 #include <cstring> 12 #include <ctime> 13 #include <iomanip> 14 #include <iostream> 15 #include <map> 16 #include <queue> 17 #include <string> 18 #include <set> 19 #include <vector> 20 21 using namespace std; 22 #define pr(x) cout << #x << " = " << x << " " 23 #define prln(x) cout << #x << " = " << x << endl 24 const int N = 2e3 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7; 25 const int OFF = 3000; 26 27 int n, a[N], b[N]; 28 int f[2][OFF * 2 + 5]; //前i个任务 且A比B多用时j 的最小时间花费 29 30 void getMin(int& x, int y) { 31 if(x > y) x = y; 32 } 33 34 int main() { 35 #ifdef LOCAL 36 freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin); 37 // freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout); 38 #endif 39 ios_base::sync_with_stdio(0); 40 clock_t _ = clock(); 41 42 int t; scanf("%d", &t); 43 while(t--) { 44 scanf("%d", &n); 45 for(int i = 1; i <= n; ++i) scanf("%d%d", a + i, b + i); 46 int z = 0; 47 memset(f[z], 0x3f, sizeof f[z]); 48 f[z][OFF] = 0; 49 for(int i = 1; i <= n; ++i) { 50 memset(f[!z], 0x3f, sizeof f[!z]); 51 for(int j = -OFF; j <= OFF; ++j) { 52 if(j > 0) { 53 getMin(f[!z][a[i] + OFF], f[z][j + OFF] + a[i]); //给A B只能等 54 getMin(f[!z][j + OFF - b[i]], f[z][j + OFF] + max(0, b[i] - j)); 55 } else { 56 getMin(f[!z][-b[i] + OFF], f[z][j + OFF] + b[i]); //给B A只能等 57 getMin(f[!z][j + OFF + a[i]], f[z][j + OFF] + max(0, a[i] + j)); 58 } 59 } 60 z = !z; 61 } 62 int ans = *min_element(f[z], f[z] + 2 * OFF + 1); 63 printf("%d\n", ans); 64 } 65 66 #ifdef LOCAL 67 printf("\nTime cost: %.2fs\n", 1.0 * (clock() - _) / CLOCKS_PER_SEC); 68 #endif 69 return 0; 70 }
