HDU 1964 Pipes

HDU_1964

    这个题目只需要把求回路数量的dp方程改写成取最优解的dp方程即可。

    更多和插头dp相关的题目可以参考胡浩的博客：http://www.notonlysuccess.com/index.php/plug-dp-complete/。

#include<stdio.h>
#include<string.h>
#define MAXD 15
#define HASH 30007
#define SIZE 1000010
#define INF 0x3f3f3f3f
int N, M, code[MAXD], ch[MAXD], maze[MAXD][MAXD], rcost[MAXD][MAXD], dcost[MAXD][MAXD];
char b[50];
struct Hashmap
{int head[HASH], next[SIZE], state[SIZE], f[SIZE], size;void init(){memset(head, -1, sizeof(head));size = 0;}void push(int st, int ans){int i, h = st % HASH;for(i = head[h]; i != -1; i = next[i])if(state[i] == st){if(ans < f[i])f[i] = ans;return ;}state[size] = st, f[size] = ans;next[size] = head[h];head[h] = size ++;}
}hm[2];
void decode(int *code, int m, int st)
{int i;for(i = m; i >= 0; i --){code[i] = st & 7;st >>= 3;}
}
int encode(int *code, int m)
{int i, cnt = 0, st = 0;memset(ch, -1, sizeof(ch));ch[0] = 0;for(i = 0; i <= m; i ++){if(ch[code[i]] == -1)ch[code[i]] = ++ cnt;code[i] = ch[code[i]];st <<= 3;st |= code[i];}return st;
}
void init()
{int i, j, k;gets(b), sscanf(b, "%d%d", &N, &M);memset(maze, 0, sizeof(maze));for(i = 1; i <= N; i ++)for(j = 1; j <= M; j ++)maze[i][j] = 1;gets(b);for(i = 1; i < 2 * N; i ++){gets(b);if(i & 1){for(j = 1, k = 2; k < 2 * M; j ++, k += 2)rcost[(i + 1) >> 1][j] = b[k] - '0';}else{for(j = 1, k = 1; k < 2 * M; j ++, k += 2)dcost[i >> 1][j] = b[k] - '0';}}gets(b);
}
void shift(int *code, int m)
{int i;for(i = m; i > 0; i --)code[i] = code[i - 1];code[0] = 0;
}
void dpblank(int i, int j, int cur)
{int k, left, up, t;for(k = 0; k < hm[cur].size; k ++){decode(code, M, hm[cur].state[k]);left = code[j - 1], up = code[j];if(left && up){if(left == up){if(i == N && j == M){code[j - 1] = code[j] = 0;shift(code, M);hm[cur ^ 1].push(encode(code, M), hm[cur].f[k]);}}else{code[j - 1] = code[j] = 0;for(t = 0; t <= M; t ++)if(code[t] == up)code[t] = left;if(j == M)shift(code, M);hm[cur ^ 1].push(encode(code, M), hm[cur].f[k]);}}else if(left || up){if(maze[i][j + 1]){code[j - 1] = 0, code[j] = left + up;hm[cur ^ 1].push(encode(code, M), hm[cur].f[k] + rcost[i][j]);}if(maze[i + 1][j]){code[j - 1] = left + up, code[j] = 0;if(j == M)shift(code, M);hm[cur ^ 1].push(encode(code, M), hm[cur].f[k] + dcost[i][j]);}}else{if(maze[i][j + 1] && maze[i + 1][j]){code[j - 1] = code[j] = 13;hm[cur ^ 1].push(encode(code, M), hm[cur].f[k] + rcost[i][j] + dcost[i][j]);}}}
}
void solve()
{int i, j, cur = 0, ans = INF;hm[cur].init();hm[cur].push(0, 0);for(i = 1; i <= N; i ++)for(j = 1; j <= M; j ++){hm[cur ^ 1].init();dpblank(i, j, cur);cur ^= 1;}for(i = 0; i < hm[cur].size; i ++)if(hm[cur].f[i] < ans)ans = hm[cur].f[i];printf("%d\n", ans);
}
int main()
{int t;gets(b), sscanf(b, "%d", &t);while(t --){init();solve();}return 0;
}