Optimal matrix chain multiplication

The main large interval into the inter-district ...

First small interval evaluation ...

State transfer equation F (i,j) = min{F (i,k) + f (k+1,j) + p[i-1]p[k]p[j]};

POJ 1651 http://poj.org/problem?id=1651

#include <iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<cstdlib>Const intMAXN = -+Ten;Const DoubleESP = 10e-8;Const DoublePi = Atan (1.0) *4;Const intINF =0xFFFFFFF;Const intMOD =10000007;using namespacestd;intN;intDP[MAXN][MAXN];intC[MAXN];voidmultiplication () { for(inti =0; I < n;i++) {Dp[i][i]=0; }     for(intL =2; l < n;l++){         for(inti =1; I < n-l+1; i++){            intj = i+l-1; DP[I][J]=INF;  for(intK = I;k < j;k++){                intTMP = Dp[i][k] + dp[k+1][J] + c[i-1] * C[k] *C[j]; DP[I][J]=min (dp[i][j],tmp); }        }    }}intMain () {//freopen ("Input.txt", "R", stdin);scanf"%d",&N);  for(inti =0; I < n;i++) {scanf ("%d",&C[i]);    } multiplication (); printf ("%d\n", dp[1][n-1]); return 0;}

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POJ 1179 http://poj.org/problem?id=1179

Approximate test instructions:

Give you a ring, the point is the number, the edge is an operation, you break one of the edges, and then you can combine the two connected points through the side of the operation into a point, to find the maximum value

To solve the problem, enumerate the broken points and multiply the matrices.

The ... 1 I can't copy the code ... Copying the code is not working ... Orz ... http://m.blog.csdn.net/blog/u012962816/26822959

The ... 2 Http://www.tuicool.com/articles/Jj2Avi

When the first one is stupid enough to copy the wrong, refer to the 2

Then I wrote it from the new ...

#include <iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<cstdlib>Const intMAXN = -+Ten;Const DoubleESP = 10e-8;Const DoublePi = Atan (1.0) *4;Const intINF =0xFFFFFF;Const intMOD =10000007;using namespacestd;CharEDGE[MAXN];intVET[MAXN];intdp1[maxn/2][maxn/2],dp2[maxn/2][maxn/2];intFIN[MAXN];intN;intMain () {//freopen ("Input.txt", "R", stdin);    intAns =-INF; scanf ("%d",&N);  for(inti =0; I < n;i++) {GetChar (); scanf ("%c%d",&edge[i],&Vet[i]); }     for(intt =0; t < n;t++) {//from the T position as the starting point         for(inti =0; i < n;i++) {//initialization of the matrixDp1[i][i] = vet[(t+i)%N]; Dp2[i][i]=Dp1[i][i]; }           for(intL =2; l <= n;l++) {//Length of J-i             for(inti =0; I < n-l+1; i++){                 intMAXV =-INF; intMINV =INF; intj = i+l-1;  for(intK = I;k < j;k++){                    if(edge[(k +1+t)%N] = ='T'){//It 's going to change here, too.MAXV = Max (Maxv,dp1[i][k] + dp1[(k +1)%N]                        [j]); MINV= Min (Minv,dp2[i][k] + dp2[(k +1)%N]                    [j]); }                    Else{MAXV= MAX (maxv,dp1[i][k] * dp1[(k +1)%N]                        [j]); MAXV= MAX (maxv,dp2[i][k] * dp2[(k +1)%n][j]);//negative Negative positiveMINV = min (minv,dp1[i][k] * dp2[(k +1)%n][j]);//positive and negative multiplicationMINV = min (minv,dp2[i][k] * dp2[(k +1)%N]                        [j]); MINV= MIN (minv,dp2[i][k] * dp1[(k +1)%n][j]);//positive and negative multiplication}} Dp1[i][j]= MAXV;//to assign a valueDP2[I][J] =MINV; }} ans= Max (ans,dp1[0][n-1]); Fin[t]= dp1[0][n-1]; } printf ("%d\n", ans);  for(inti =0; I < n;i++){        if(Fin[i] = =ans) {printf ("%d", i+1); }} printf ("\ n"); return 0;}

UVA 10003 http://uva.onlinejudge.org/index.php?option=com_onlinejudge&itemid=8&page=show_problem& category=847&problem=944&mosmsg=submission+received+with+id+15274403

D (I,J) is the best cost for i-j, then D (i,j)-min{D (i,k) + D (k,j) + a[j]-a[i]};

First small interval evaluation ...

#include <iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<cstdlib>Const intMAXN = ++Ten;Const DoubleESP = 10e-8;Const DoublePi = Atan (1.0) *4;Const intINF =0xFFFFFF;Const intMOD =10000007;using namespacestd;intA[MAXN];intDP[MAXN][MAXN];intMain () {//freopen ("Input.txt", "R", stdin);    intLen;  while(~SCANF ("%d", &len) &&Len) {        intN; scanf ("%d",&N);  for(inti =1; I <= n;i++) {scanf ("%d",&A[i]); } memset (DP,0,sizeof(DP)); a[0] =0; A[n+1] =Len;  for(intL =2; l <= n+1; l++){             for(inti =0; I+l <= n+1; i++){                intj = i+l; DP[I][J]=INF;  for(intK = i+1; k < j;k++){                    intTMP = Dp[i][k] + dp[k][j] + a[j]-A[i]; if(TMP <Dp[i][j]) {Dp[i][j]=tmp; } }}} printf ("The minimum cutting is%d.\n", dp[0][n+1]); }    return 0;}

Optimal matrix chain multiplication