Noi question Bank/2.6 Dynamic planning of basic algorithms-7614: Minimum tolls

Total time limit:

1000ms

Memory Limit:

65536kB

Describe

A businessman walks through a n*n square grid to attend a very important business event. He's going to go from the top left corner of the grid and out the bottom right corner. It takes 1 units of time to cross 1 squares in the middle of each one. The businessman must cross out at (2n-1) a unit time. And in the middle of every small box, you need to pay a certain fee.

The businessman expected to cross out at a minimum cost within the stipulated time. How much do I need at least?

Note: You cannot cross the small squares diagonally (that is, you can only move up and down in four directions and not leave the grid).

Input

The

first line is an integer representing the width of the square n (1 <= n < 100);
The next n rows, each row n an integer not greater than 100, are the cost of each small square on the grid.

Output

at least the required cost.

Sample input

Sample output

109

Tips

in the sample, the minimum value is 109=1+2+5+7+9+12+19+21+33.

Source

Yuanpei-from WHF

Topic links

Cost Flow

#include <cstdio>#include<queue>#defineM 100000#defineN 105#defineINF 0x3f3f3f3fusing namespacestd;BOOLInq[n];intN,s,t,cnt=1, fx[5]= {1,-1,0,0},fy[5]= {0,0,-1,1},a[n][n],to[m<<1],dis[m],fa[m],head[m],came[m<<1],nxt[m<<1],flow[m<<1],cost[m<<1];voidInsintUintVintWintf) {nxt[++cnt]=head[u];to[cnt]=v;flow[cnt]=w;cost[cnt]=f;head[u]=CNT; nxt[++cnt]=head[v];to[cnt]=u;flow[cnt]=0; cost[cnt]=-f;head[v]=CNT;}BOOLSPFA () { for(intI=s; i<=t; ++i) dis[i]=inf,inq[i]=false, came[i]=inf; Queue<int>Q; Dis[s]=0;    Q.push (S);  for(intu;!Q.empty ();) {u=Q.front ();        Q.pop (); Inq[u]=false;  for(intI=head[u]; I I=Nxt[i]) {            intv=To[i]; if(dis[v]>dis[u]+cost[i]&&Flow[i]) {Dis[v]=dis[u]+Cost[i]; FA[V]=i; CAME[V]=min (came[u],flow[i]); if(!Inq[v]) {Inq[v]=true;                Q.push (v); }            }        }    }    returndis[t]<inf;}intMainintargcChar*argv[]) {scanf ("%d",&N); S=0; T=n*n*2+1;  for(intI=1; i<=n; ++i) for(intj=1; j<=n; ++j) scanf ("%d",&A[i][j]); Ins (S,1, inf,a[1][1]); INS (n*n,t,inf,0);  for(intI=1; i<=n; ++i) { for(intj=1; j<=n; ++j) {ins ((I-1) *n+j, (I-1) *n+j+n*n,1,0);  for(intk=0; k<4; ++k) {intti=i+fx[k],tj=j+Fy[k]; if(ti<1|| ti>n| | tj<1|| Tj>n)Continue; Ins ((I-1) *n+j+n*n, (ti-1) *N+TJ,1, A[TI][TJ]); }        }    }    intans=0;  while(SPFA ()) {intt=Came[t];  for(intl=1; I!=s; i=to[fa[i]^1]) {Flow[fa[i]]-=T; Flow[fa[i]^1]+=T; } ans+=t*Dis[t]; } printf ("%d\n", ans); return 0;}

Dp

#include <iostream>using namespacestd;intn,a[118][118],f[118][118];intMain () {CIN>>N;  for(intI=1; i<=n; i++)         for(intj=1; j<=n; J + +) {cin>>A[i][j]; if(j==1) F[i][j]=f[i-1][j]+A[i][j]; Else if(i==1) F[i][j]=f[i][j-1]+A[i][j]; ElseF[i][j]=min (f[i-1][j]+a[i][j],f[i][j-1]+A[i][j]); } cout<<F[n][n]; return 0;}

Noi question Bank/2.6 Dynamic planning of basic algorithms-7614: Minimum tolls