Magazine Delivery 动态规划 记忆式递归实现

三辆车在三个点作为状态，位置依次增大作为状态标识，状态函数标识当前状态距离最终状态还需要行驶的距离。

#include<iostream>

using namespace std;

int n;

int graph[31][31];

int flag[31][31][31];

int D(int a,int b,int c){

int tmp = 99999;

if(c == n) return 0;

if(!flag[b][c][c+1])flag[b][c][c+1] = D(b,c,c+1);

if(!flag[a][c][c+1])flag[a][c][c+1] = D(a,c,c+1);

if(!flag[a][b][c+1])flag[a][b][c+1] = D(a,b,c+1);

if(tmp>flag[b][c][c+1]+graph[a][c+1])tmp = flag[b][c][c+1]+graph[a][c+1];

if(tmp>flag[a][c][c+1]+graph[b][c+1])tmp = flag[a][c][c+1]+graph[b][c+1];

if(tmp>flag[a][b][c+1]+graph[c][c+1])tmp = flag[a][b][c+1]+graph[c][c+1];

return tmp;

}

int main(){

int i,j,t;

cin >> t;

while(t--){

cin >> n;

memset(graph,0,sizeof(int)*31*31);

memset(flag,0,sizeof(int)*31*31*31);

for(i=1;i<=n-1;i++){

for(j=i+1;j<=n;j++){

cin >> graph[i][j];

graph[j][i] = graph[i][j];

}

}

cout << D(1,1,1)<<endl;

}

}