Gerrymandering
闅撅紒O(n^3)鐨刣p錛岄檷涓烘帴榪慜(n^2)鎵嶈兘榪囥?br>
涓嶉毦鎯沖埌O(n^3)鐨刣p錛宒p[k][t]浠ｈ〃鍓峩涓猺iding鍚堝茍鎴恡鍧楋紙涔熷氨鏄鍚堝茍k-t嬈★級錛屾墍鑳藉彇寰楃殑鏈澶у鹼紙璧㈢殑riding涓暟鍑忓幓杈撶殑riding涓暟錛夈傚畾涔墂[i][j]浠ｈ〃i鑷砵鍚堝茍涓璧峰悗鐨勮緭璧㈢姸鎬侊紝1鎴?1銆?br>1銆乨p[0][0] = 0錛岀姸鎬佽漿縐繪柟紼媎p[k][t] = max(dp[k'][t-1] + win[k'+1][k]), 0<=k'<k銆?br>2銆佺洰鏍囧垯鏄痬ax(t) 浣垮緱dp[n][t] > 0錛岃緭鍑虹瓟妗坣-t銆?br>杞Щ鐨勫鏉傚害鏄痮(n)錛岃繖鏍鋒垜浠殑澶嶆潅搴﹀氨鏄疧(n^3)錛屽繀欏婚檷浣庡鏉傚害銆?br>
鎯蟲硶錛氭垜浠眰dp[k][t]鏃訛紝閬嶅巻浜嗕竴嬈p[k'][t-1]錛屽綋鎴戜滑鍐嶆眰dp[k+1][t]鏃訛紝榪橀渶瑕佸啀閬嶅巻涓嬈p[k'][t-1]錛屽彧涓嶈繃榪欎釜鏃跺檏'鑼冨洿澧炲姞浜?銆傞偅鎴戜滑鑳藉惁鍦ㄦ眰dp[k][t]閬嶅巻dp[k'][t-1]鏃舵妸榪欎簺鏈夌敤淇℃伅璁板綍涓嬫潵渚涙眰dp[k+1][t]鏃剁敤鍛紵
鐪嬬姸鎬佽漿縐繪柟紼嬶細dp[k][t] = max(dp[k'][t-1] + win[k'+1][k])
闂錛氬彈win[k'+1][k]褰卞搷錛宒p[k'][t-1] 鐨勬渶澶у鹼紝鍔犱笂win[k'+1][k]鍚庡茍涓嶄竴瀹氭槸dp[k][t]鐨勬渶澶у?br>浣嗘槸錛氫粩緇嗙湅浼氬彂鐜幫紝win[k'+1][k]浠呮湁涓ょ鍙栧鹼紝1鍜?1,鑰屼笖褰搆'鍙樺寲鏃禿p[k'][t-1]鐨勫肩浉宸?銆?銆?
榪欐牱鐨勮瘽錛屾柟紼嬪彲浠ュ啓鎴愶細dp[k][t] = maxdp + win[k''+1][k]錛岃繖閲宮axdp = max(dp[k'][t-1])錛宬''鏈変竴涓檺鍒訛紝灝辨槸錛宒p[k''][t-1]蹇呴』絳変簬maxdp銆?br>涓鏃﹁漿鎹㈡垚榪欑被鏂圭▼鍚庯紝dp[k'][t-1]鐨勪俊鎭氨鍙互閲嶅鍒╃敤錛岀姸鎬佽漿縐葷殑澶嶆潅搴﹀ぇ澶ч檷浣庛?br>
瀹炵幇錛氳褰昺axdp錛岀劧鍚庡緩涓涓猯ist錛岃褰曟弧瓚砫p[k''][t-1] = maxdp鐨刱'’錛屽疄鏃舵洿鏂發ist錛屾瘡嬈″彧闇浠巐ist閲岃漿縐葷姸鎬佸氨琛屻?br>浼樺寲錛氫竴鏃︿粠list閲屾壘鍒頒竴涓猭''錛屾弧瓚硍in[k''+1][k]=1錛岄偅涔坉p[k][t] = maxdp + 1錛宒p[k][t]鐨勮綆楀氨鍙┈涓婄粨鏉熴?br>
1 //POJ_3351 锝?nbsp;alpc02
2 #include <algorithm>
3 using namespace std;
4
5 const int INF = 12345678;
6 const int N = 1010;
7 const int P = 12;
8
9 int n, p;
10 int v[N][P];
11 int win[N][N];
12 int dp[N][N];
13
14 int Max(int a,int b){return a>b ? a:b;}
15 void input() {
16 scanf("%d", &n);
17 scanf("%d", &p);
18 int i, j, k;
19 memset(v[0], 0, sizeof(v[0]));
20 for(i=1; i<=n; i++)
21 for(j=1; j<=p; j++) {
22 scanf("%d", &k);
23 v[i][j] = v[i-1][j] + k;
24 }
25 }
26 void init() {
27 int i, j, k, vote;
28 for(i=1; i<=n; i++)
29 for(j=i; j<=n; j++) {
30 vote = v[j][1] - v[i-1][1];
31 for(k=2; k<=p; k++)
32 if(v[j][k] - v[i-1][k] >= vote)
33 break;
34 if(k > p) win[i][j] = 1;
35 else win[i][j] = -1;
36 }
37 }
38 int solve() {
39 init();
40
41 int i, j, k, t;
42 int list[N], top, mx;
43 dp[0][0] = 0;
44 for(j=1; j<=n; j++) {
45 top = 0;
46 mx = -INF;
47 for(i=j; i<=n; i++) {
48 if(dp[i-1][j-1] > mx) {
49 mx = dp[i-1][j-1];
50 list[0] = i-1;
51 top = 1;
52 }
53 else if(dp[i-1][j-1] == mx)
54 list[top++] = i-1;
55 for(t=0; t<top-1; t++)
56 if(win[list[t]+1][i] == 1)
57 break;
58 dp[i][j] = mx + win[list[t]+1][i];
59 }
60 }
61
62 int ans;
63 for(ans=n; ans>=1; ans--)
64 if(dp[n][ans] > 0)
65 return n-ans;
66 return -1;
67 }
68 int main() {
69 input();
70 printf("%d\n", solve());
71 return 0;
72 }
2 #include <algorithm>
3 using namespace std;
4
5 const int INF = 12345678;
6 const int N = 1010;
7 const int P = 12;
8
9 int n, p;
10 int v[N][P];
11 int win[N][N];
12 int dp[N][N];
13
14 int Max(int a,int b){return a>b ? a:b;}
15 void input() {
16 scanf("%d", &n);
17 scanf("%d", &p);
18 int i, j, k;
19 memset(v[0], 0, sizeof(v[0]));
20 for(i=1; i<=n; i++)
21 for(j=1; j<=p; j++) {
22 scanf("%d", &k);
23 v[i][j] = v[i-1][j] + k;
24 }
25 }
26 void init() {
27 int i, j, k, vote;
28 for(i=1; i<=n; i++)
29 for(j=i; j<=n; j++) {
30 vote = v[j][1] - v[i-1][1];
31 for(k=2; k<=p; k++)
32 if(v[j][k] - v[i-1][k] >= vote)
33 break;
34 if(k > p) win[i][j] = 1;
35 else win[i][j] = -1;
36 }
37 }
38 int solve() {
39 init();
40
41 int i, j, k, t;
42 int list[N], top, mx;
43 dp[0][0] = 0;
44 for(j=1; j<=n; j++) {
45 top = 0;
46 mx = -INF;
47 for(i=j; i<=n; i++) {
48 if(dp[i-1][j-1] > mx) {
49 mx = dp[i-1][j-1];
50 list[0] = i-1;
51 top = 1;
52 }
53 else if(dp[i-1][j-1] == mx)
54 list[top++] = i-1;
55 for(t=0; t<top-1; t++)
56 if(win[list[t]+1][i] == 1)
57 break;
58 dp[i][j] = mx + win[list[t]+1][i];
59 }
60 }
61
62 int ans;
63 for(ans=n; ans>=1; ans--)
64 if(dp[n][ans] > 0)
65 return n-ans;
66 return -1;
67 }
68 int main() {
69 input();
70 printf("%d\n", solve());
71 return 0;
72 }
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