【DP】 HDU 5092 Seam Carving

来源：互联网发布：数据分析的具体流程编辑：程序博客网时间：2024/05/29 13:46

从上到下找到一条路要路上的值最小

如果有多条路径输出最靠右边的

#include <cstdio>#include <cstring>#include <cstdlib>#include <string>#include <iostream>#include <algorithm>#include <sstream>#include <cmath>using namespace std;#include <queue>#include <stack>#include <vector>#include <deque>#define cler(arr, val)    memset(arr, val, sizeof(arr))typedef long long  LL;const int MAXN = 100200;const int MAXM = 6000010;const int INF = 0x3f3f3f3f;const int mod = 1000000007;LL a[133][133],dp[133][133],pre[133][133];vector<int>point;int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);    // freopen("out.txt", "w", stdout);#endif    int t,n,m,cas=1;    cin>>t;    while(t--)    {        cin>>n>>m;        cler(a,0);cler(pre,0);        cler(dp,0);        for(int i=0;i<=n;i++)            for(int j=0;j<=m+2;j++)                dp[i][j]=(LL)INF;        for(int i=1;i<=n;i++)            for(int j=1;j<=m;j++)                scanf("%I64d",&a[i][j]);        for(int i=n;i>=1;i--)        {            for(int j=1;j<=m;j++)            {                for(int k=-1;k<=1;k++)                {                    if(dp[i][j]>=dp[i+1][j+k]+a[i][j])                    {                        pre[i][j]=j+k;                        dp[i][j]=dp[i+1][j+k]+a[i][j];                    }                }            }        }        printf("Case %d\n",cas++);        LL ans=INF;        point.clear();        for(int i=1;i<=m;i++)        {            if(ans>dp[1][i])            {                point.clear();                ans=dp[1][i];                point.push_back(i);            }            else if(ans==dp[1][i])                point.push_back(i);        }        LL num,sum=0;        for(int j=0;j<point.size();j++)        {            LL p=point[j];            LL cc=0;            for(int i=1;i<=n;i++)            {                cc+=p;                p=pre[i][p];            }            if(cc>sum)                sum=cc,num=j;        }        LL p=point[num];        for(int i=1;i<=n;i++)        {            printf("%I64d%c",p,i==n?'\n':' ');            p=pre[i][p];        }        //cout<<ans<<endl;    }    return 0;}

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