标签:ons map isa span iii 注意 tmp ios hdu
理解:
http://blog.renren.com/share/263498909/1064362501
http://www.cnblogs.com/ronaflx/archive/2011/03/30/1999764.htmlhttp://yomean.blog.163.com/blog/static/189420225201272864127683/
http://www.cnblogs.com/zxndgv/archive/2011/08/02/2125242.html
题目总结:
http://www.cnblogs.com/ronaflx/archive/2011/03/30/1999764.html
下摘自:http://www.cnblogs.com/zxndgv/archive/2011/08/02/2125242.html
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最有代价用d[i,j]表示
d[i,j]=min{d[i,k-1]+d[k+1,j]}+w[i,j]
当中w[i,j]=sum[i,j]
四边形不等式
w[a,c]+w[b,d]<=w[b,c]+w[a,d](a<b<c<d) 就称其满足凸四边形不等式
决策单调性
w[i,j]<=w[i‘,j‘] ([i,j]属于[i‘,j‘]) 既 i‘<=i<j<=j‘
于是有下面三个定理
定理一: 假设w同一时候满足四边形不等式 和 决策单调性 ,则d也满足四边形不等式
定理二:当定理一的条件满足时,让d[i,j]取最小值的k为K[i,j],则K[i,j-1]<=K[i,j]<=K[i+1,j]
定理三:w为凸当且仅当w[i,j]+w[i+1,j+1]<=w[i+1,j]+w[i,j+1]
由定理三知 推断w是否为凸即推断 w[i,j+1]-w[i,j]的值随着i的添加是否递减
于是求K值的时候K[i,j]仅仅和K[i+1,j] 和 K[i,j-1]有关。所以 能够以i-j递增为顺序递推各个状态值终于求得结果 将O(n^3)转为O(n^2)
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注意:
注意决策单调性顺序,既要符合决策调性,也要符合题意 (!!!!)
注意枚举顺序,依据dp方程和决策单调性方程
注意初始化,防止訪问到无效状态或没处理的状态。
dp和s边界初始化,尤其是决策的上届和下届初始化。
例题1:
石子合并问题:hdu 3506
dp[i][j] = min{dp[i][k] + dp[k + 1][j] + cost[i, j] }, i <= k <= j - 1 , cost[i][j] = sum[j] - sum[i - 1]
s[i][j - 1] <= s[i][j] <= s[i + 1][j]
//#pragma warning (disable: 4786) //#pragma comment (linker, "/STACK:16777216") //HEAD #include <cstdio> #include <ctime> #include <cstdlib> #include <cstring> #include <queue> #include <string> #include <set> #include <stack> #include <map> #include <cmath> #include <vector> #include <iostream> #include <algorithm> using namespace std; //LOOP #define FE(i, a, b) for(int i = (a); i <= (b); ++i) #define FD(i, b, a) for(int i = (b); i>= (a); --i) #define REP(i, N) for(int i = 0; i < (N); ++i) #define CLR(A,value) memset(A,value,sizeof(A)) #define CPY(a, b) memcpy(a, b, sizeof(a)) #define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++) //INPUT #define RI(n) scanf("%d", &n) #define RII(n, m) scanf("%d%d", &n, &m) #define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k) #define RS(s) scanf("%s", s) //OUTPUT #define WI(n) printf("%d\n", n) #define WS(s) printf("%s\n", s) typedef long long LL; const int INF = 1000000007; const double eps = 1e-10; const int maxn = 2010; int dp[maxn][maxn], s[maxn][maxn]; int w[maxn][maxn]; int n, m; int val[maxn]; int sum[maxn]; ///求dp最小值 ///枚举 区间 由小到大 void solve() { // memset(dp, 0, sizeof(dp));///初始化无效值 FE(i, 1, 2 * n) { dp[i][i] = 0; s[i][i] = i;/// 初始化决策下届,为0 } FE(len, 2, n) { for (int i = 2 * n - len; i >= 1; i--) { int j = i + len - 1; dp[i][j] = INF; int a = s[i][j - 1], b = s[i + 1][j]; int cost = sum[j] - sum[i - 1]; for (int k = a; k <= b; k++) { if (dp[i][j] > dp[i][k] + dp[k + 1][j] + cost) { dp[i][j] = dp[i][k] + dp[k + 1][j] + cost; s[i][j] = k; } } } } } int main () { while (~RI(n)) { FE(i, 1, n) RI(val[i]), val[i + n] = val[i];; FE(i, 1, 2 * n) sum[i] = sum[i - 1] + val[i]; // pre(); solve(); int ans = INF; FE(i, 1, n) { if (ans > dp[i][n + i - 1]) ans = dp[i][n + i - 1]; } printf("%d\n", ans); } return 0; }
1:注意此法的 i 和 j 顺序与寻常不同
此时决策区间为:s[i - 1][j] <= s[i][j] <= s[i][j + 1] (!!!)
详细见凝视:
//#pragma warning (disable: 4786) //#pragma comment (linker, "/STACK:16777216") //HEAD #include <cstdio> #include <ctime> #include <cstdlib> #include <cstring> #include <queue> #include <string> #include <set> #include <stack> #include <map> #include <cmath> #include <vector> #include <iostream> #include <algorithm> using namespace std; //LOOP #define FE(i, a, b) for(int i = (a); i <= (b); ++i) #define FD(i, b, a) for(int i = (b); i>= (a); --i) #define REP(i, N) for(int i = 0; i < (N); ++i) #define CLR(A,value) memset(A,value,sizeof(A)) #define CPY(a, b) memcpy(a, b, sizeof(a)) #define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++) //INPUT #define RI(n) scanf("%d", &n) #define RII(n, m) scanf("%d%d", &n, &m) #define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k) #define RS(s) scanf("%s", s) //OUTPUT #define WI(n) printf("%d\n", n) #define WS(s) printf("%s\n", s) typedef long long LL; const int INF = 1000000007; const double eps = 1e-10; const int maxn= 1000010; int dp[33][333], s[33][333]; int w[333][333]; int n, m; int val[333]; int sum[333]; ///dp[i][j] = min{dp[i - 1][k] + w[k + 1][j]}, i - 1 <= k <= j - 1 ///一般要求 i <= j (!!) ///s[i - 1][j] <= s[i][j] <= s[i][j + 1] (!!
) ///注意决策单调性顺序,既要符合决策调性,也要符合题意 (!!!!) ///注意枚举顺序,依据dp方程和决策单调性方程 ///注意初始化,防止訪问到无效状态或没处理的状态。dp和s边界初始化。尤其是决策的上届和下届初始化。 void pre() { for(int i = 1; i <= n; i ++) //这里有一个递推公式能够进行预处理 { w[i][i] = 0; for(int j = i + 1; j <= n; j ++) { int mid = (j + i) >> 1; w[i][j] = w[i][j - 1] + val[j] - val[mid]; // int x = sum[j] - sum[mid] - val[mid] * (j - mid); // x += val[mid] * (mid - i) - (sum[mid - 1] - sum[i - 1]); } } } ///求dp最小值 ///枚举i从小到大 ///再枚举j从大到小 void solve() { memset(dp, 0, sizeof(dp));///初始化无效值 FE(i, 1, n) { dp[1][i] = w[1][i]; s[1][i] = 0;/// 初始化决策下届,为0 } FE(i, 2, m) { //s[1][i] = 0; s[i][n + 1] = n;///初始化决策上届 for (int j = n; j >= i; j--) { int tmp = dp[i][j] = INF;///初始化最优值 int a = s[i - 1][j], b = s[i][j + 1]; //a = max(a, i - 1); b = min(b, j - 1); // i - 1 <= k <= j - 1 for (int k = a; k <= b; k++) ///保证枚举到的都是有效状态,且都已计算过 { if (tmp > dp[i - 1][k] + w[k + 1][j]) { tmp = dp[i - 1][k] + w[k + 1][j]; s[i][j] = k; } } dp[i][j] = tmp; } } } int main () { while (~RII(n, m)) { FE(i, 1, n) RI(val[i]), sum[i] = sum[i - 1] + val[i]; pre(); solve(); printf("%d\n", dp[m][n]); } return 0; }
此时决策区间为:s[i][j - 1] <= s[i][j] <= s[i + 1][j] (!!!)
//#pragma warning (disable: 4786) //#pragma comment (linker, "/STACK:16777216") //HEAD #include <cstdio> #include <ctime> #include <cstdlib> #include <cstring> #include <queue> #include <string> #include <set> #include <stack> #include <map> #include <cmath> #include <vector> #include <iostream> #include <algorithm> using namespace std; //LOOP #define FE(i, a, b) for(int i = (a); i <= (b); ++i) #define FD(i, b, a) for(int i = (b); i>= (a); --i) #define REP(i, N) for(int i = 0; i < (N); ++i) #define CLR(A,value) memset(A,value,sizeof(A)) #define CPY(a, b) memcpy(a, b, sizeof(a)) #define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++) //INPUT #define RI(n) scanf("%d", &n) #define RII(n, m) scanf("%d%d", &n, &m) #define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k) #define RS(s) scanf("%s", s) //OUTPUT #define WI(n) printf("%d\n", n) #define WS(s) printf("%s\n", s) typedef long long LL; const int INF = 1000000007; const double eps = 1e-10; const int maxn= 1000010; int dp[333][33], s[333][33]; int w[333][333]; int n, m; int val[333]; int sum[333]; ///dp[i][j] = min{dp[k][j - 1] + w[k + 1][j]}, j - 1 <= k <= i - 1 ///此处i>=j ///s[i][j - 1] <= s[i][j] <= s[i + 1][j] (!!) ///注意决策单调性顺序,既要符合决策调性,也要符合题意 (!!!!) ///注意枚举顺序。依据dp方程和决策单调性方程 ///注意初始化。防止訪问到无效状态或没处理的状态。
dp和s边界初始化,尤其是决策的上届和下届初始化。 void pre() { for(int i = 1; i <= n; i ++) //这里有一个递推公式能够进行预处理 { w[i][i] = 0; for(int j = i + 1; j <= n; j ++) { int mid = (j + i) >> 1; w[i][j] = w[i][j - 1] + val[j] - val[mid]; } } } ///求dp最小值 ///枚举i从小到大 ///再枚举j从大到小 void solve() { memset(dp, 0, sizeof(dp));///初始化无效值 FE(i, 1, n) { dp[i][1] = w[1][i]; s[i][1] = 0;/// 初始化决策下届,为0 } FE(i, 2, m) { //s[i][1] = 0; s[n + 1][i] = n;///初始化决策上届 for (int j = n; j >= i; j--) { int tmp = dp[j][i] = INF;///初始化最优值 int a = s[j][i - 1], b = s[j + 1][i]; //a = max(a, i - 1); b = min(b, j - 1); for (int k = a; k <= b; k++) ///保证枚举到的都是有效状态,且都已计算过 { if (tmp > dp[k][i - 1] + w[k + 1][j]) { tmp = dp[k][i - 1] + w[k + 1][j]; s[j][i] = k; } } dp[j][i] = tmp; } } } int main () { while (~RII(n, m)) { FE(i, 1, n) RI(val[i]); pre(); solve(); printf("%d\n", dp[n][m]); } return 0; }
标签:ons map isa span iii 注意 tmp ios hdu
原文地址:http://www.cnblogs.com/cynchanpin/p/6727207.html