标签：des style blog class code c
Time Limit: 2000/1000 MS
(Java/Others) Memory Limit: 32768/32768 K
(Java/Others)
Total Submission(s): 1453 Accepted
Submission(s): 500
3 4 1 -1 1 0 2 -2 4 2 3 5 1 -90
1 (0 ,0,8) --> 1 8
3 (-2,1,9) --> 3 9
6 (3,6,11) --> 6 11
在此只算两列，后面的和前面一样
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#include <iostream>#include <cstdio>#include <cstring>#define M 105#define INF 0x0f0f0f0fusing
namespace std;int c[M][M]; // 存每个方格中有多少钱币int dp[M][M]; // 当前位置当前状态下最多有多少钱币void
clr(int
s[][M]){ for(int
i = 0; i < M; i++){ for(int
j = 0; j < M; j++){ s[i][j] = -INF; } }}inline
int max(int
a,int b){return
a>b?a:b;}int
main(){ int
t,n,m; scanf("%d",&t); for(int
cas = 1; cas <= t; cas++){ clr(c); clr(dp); scanf("%d%d",&n,&m); for(int
i = 1; i <= n; i++){ for(int
j = 1; j <= m; j++){ scanf("%d",&c[i][j]); } } dp[0][1] = 0; for(int
i = 1; i <= n; i++){ //计算第一列dp值 dp[i][1] = dp[i-1][1] + c[i][1]; } int
x,y; for(int
j = 2; j <= m; j++){ for(int
i = 1; i <= n; i++){ x = dp[i][j-1] + c[i][j]; //单独算第一个 y = x; dp[i][j] = max(dp[i][j],x); for(int
k = i-1; k >= 1; k--){ //往上计算 x = x + c[k][j]; dp[k][j] = max(dp[k][j],x); } for(int
k = i+1; k <= n; k++){ //往下计算 y = y + c[k][j]; dp[k][j] = max(dp[k][j],y); } } } printf("Case #%d:\n%d\n",cas,dp[1][m]); } return
0;} |
2014年百度之星程序设计大赛 - 资格赛 1004 -- Labyrinth,布布扣,bubuko.com
2014年百度之星程序设计大赛 - 资格赛 1004 -- Labyrinth
标签：des style blog class code c
原文地址：http://www.cnblogs.com/ubuntu-kevin/p/3733486.html
