poj 2353 双向dp（麻烦的办公室盖章）

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题意：给定一个n*m的整数数组。现在要从第一行中的任意点移动到最后一行的任意点，要求每次只能移动一个距离（向左右或者向下，不能向上）。求路径所包含数值之和最小的时候的路径。

思路：dp。dp[i][j] 表示从第一行走到到第i行第j列这个位置的最小代价（路径包含的所有数值之和）。显然有dp[ i ][ j ] = min(dp[ i ][ j-1 ] , dp[ i ][ j+1 ] , dp[ i-1 ][ j ])+s[ i ][ j ]。s[i][j]表示此位置的数值。那么对于每一行，需要先从左向右做一边dp，然后从右向左做一遍。同时记录路径，递归输出。

#include <stdio.h>#include <string.h>#define INF 0x3fffffff#define N 105#define M 505int dp[2][M],s[N][M],path[N][M];int n,m;void print(int n,int i){if(n > 1){if(path[n][i] == -1)print(n,i-1);else if(path[n][i] == 1)print(n,i+1);elseprint(n-1,i);}printf("%d\n",i);}int main(){freopen("a.txt","r",stdin);while(scanf("%d %d",&n,&m)!=EOF){int i,j,res;for(i = 1;i<=n;i++)for(j = 1;j<=m;j++)scanf("%d",&s[i][j]);memset(path,0,sizeof(path));dp[0][0] = dp[0][m+1] = dp[1][0] = dp[1][m+1] = INF;for(j = 1;j<=m;j++)dp[0][j] = s[1][j];for(i = 2;i<=n;i++){for(j = 1;j<=m;j++){//从左向右if(dp[1][j-1] < dp[0][j]){dp[1][j] = dp[1][j-1]+s[i][j];path[i][j] = -1;//表示从左边格子过来的}elsedp[1][j] = dp[0][j]+s[i][j];}for(j = m;j>=1;j--){//从右向左if(dp[1][j+1]+s[i][j] < dp[1][j]){dp[1][j] = dp[1][j+1]+s[i][j];path[i][j] = 1;//表示从右边格子过来的}dp[0][j] = dp[1][j];//滚动数组}}for(j = res = 1;j<=m;j++)if(dp[1][j] < dp[1][res])res = j;print(n,res);}return 0;}

其中dp数组可以连滚动数组都不用，一维数组就够了。

#include <stdio.h>#include <string.h>#define INF 0x3fffffff#define N 105#define M 505int dp[M],s[N][M],path[N][M];int n,m;void print(int n,int i){if(n > 1){if(path[n][i] == -1)print(n,i-1);else if(path[n][i] == 1)print(n,i+1);elseprint(n-1,i);}printf("%d\n",i);}int main(){freopen("a.txt","r",stdin);while(scanf("%d %d",&n,&m)!=EOF){int i,j,res;for(i = 1;i<=n;i++)for(j = 1;j<=m;j++)scanf("%d",&s[i][j]);memset(path,0,sizeof(path));dp[0] = dp[m+1] = INF;for(j = 1;j<=m;j++)dp[j] = s[1][j];for(i = 2;i<=n;i++){for(j = 1;j<=m;j++){//从左向右if(dp[j-1] < dp[j]){dp[j] = dp[j-1]+s[i][j];path[i][j] = -1;//表示从左边格子过来的}elsedp[j] = dp[j]+s[i][j];}for(j = m;j>=1;j--){//从右向左if(dp[j+1]+s[i][j] < dp[j]){dp[j] = dp[j+1]+s[i][j];path[i][j] = 1;//表示从右边格子过来的}}}for(j = res = 1;j<=m;j++)if(dp[j] < dp[res])res = j;print(n,res);}return 0;}

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