Ural 1519. Formula 1 优美的插头DP

今天早上学了插头DP的思想和最基础的应用，中午就开始敲了，岐哥说第一次写不要看别人代码，利用自己的理解一点点得写出来，这样才锻炼代码能力！于是下午慢慢地构思轮廓，一点点地敲出主体代码，其实是很磨蹭的，由于要考虑好多东西，而且昨晚2点睡的有点困，最后终于磨蹭出来了，第一次的代码搓没关系，自己写的才重要。然后果然不出我所料，调试到了晚上才A了（一个郁闷的错误）。。。A的感觉真的是爽呀，虽然搞了差不多一天。当然自己写了自己想的代码后也要把代码优化，不然队友看不懂自己代码就囧了。。。

插头DP，建议大家想学的好好看看陈丹琦的国家集训队论文，这是个优美的DP。

http://www.docin.com/p-46797997.html

 

#include <stdio.h>
#include <string.h>#define LL __int64const int mod = 10007;//   哈希表
struct HASH{int head[mod+10], E, next[80000];LL val[80000], cnt[80000];void init() {memset(head, -1, sizeof(head));E = 0;}int findhash(LL x) {return (x%mod + mod)%mod;}void add(LL x, LL sum) {int u = findhash(x);for(int i = head[u];i != -1;i = next[i]) if(val[i] == x) {cnt[i] += sum;return ;}val[E] = x;cnt[E] = sum;next[E] = head[u];head[u] = E++;}}biao1, biao2;int c[22], n, m, d[22];
//  编码
void get(LL x) {for(int i = m+1;i >= 1; i--) {c[i] = x&7;x /= 8;}
}
//  解码
LL getval() {LL ret = 0;for(int i = 1;i <= m+1; i++) {ret |= d[i];ret *= 8;}ret /= 8;return ret ;
}
//   转化成最小表示法
void change() {int vis[22];memset(vis, 0, sizeof(vis));int num = 1;for(int i = 1;i <= m+1;i ++) {if(!d[i])   continue;if(!vis[d[i]]) {vis[d[i]] = num;d[i] = num++;}else {d[i] = vis[d[i]];}}
}void fuzhi() {for(int i = 1;i <= m+1;i ++)    d[i] = c[i];
}char s[22][22];int main() {int i, j, k, l;while(scanf("%d%d", &n, &m) != -1) {for(i = 1;i <= n; i++)scanf("%s", s[i]+1);int tot = 0;for(i = 1;i <= n; i++)for(j = 1;j <= m; j++)if(s[i][j] == '.')  tot++;if(tot%2==1 || tot < 4) {puts("0");continue;}int tox  = -1, toy = -1;for(i = 1;i <= n; i++)for(j = 1;j <= m; j++) if(s[i][j] == '.') {tox = i;toy = j;}biao1.init();biao1.add(0, 1);LL ans = 0;for(i = 1;i <= n; i++){for(j = 0;j <= m; j++){biao2.init();for(l = 0;l < biao1.E; l++)  {get(biao1.val[l]);if(j == m) {for(int ii = 2;ii <= m+1; ii++) d[ii] = c[ii-1];d[1] = 0;change();LL now = getval();biao2.add(now, biao1.cnt[l]);continue;}if(c[j+1] && !c[j+2]) {  //  有左插头无上插头if(s[i][j+1] != '.')  continue;if(j+2 <= m) {fuzhi();d[j+1] = 0;d[j+2] = c[j+1];change();LL now = getval();biao2.add(now, biao1.cnt[l]);}if(i < n) {fuzhi();change();LL now = getval();biao2.add(now, biao1.cnt[l]);}}else if(!c[j+1] && c[j+2]) {  //  有上插头无左插头if(s[i][j+1] != '.')  continue;if(i < n) {fuzhi();d[j+1] = c[j+2]; d[j+2] = 0;change();LL now = getval();biao2.add(now, biao1.cnt[l]);}if(j+2 <= m) {fuzhi();change();LL now = getval();biao2.add(now, biao1.cnt[l]);}}else if(!c[j+1] && !c[j+2]) { //   左和上都无插头if(s[i][j+1] != '.') {fuzhi();change();LL now = getval();biao2.add(now, biao1.cnt[l]);continue;}if(j+2 <= m && i < n) {fuzhi();d[j+1] = d[j+2] = 13;change();LL now = getval();biao2.add(now, biao1.cnt[l]);}}else { //　　左和上都有插头　，　　要判断左和上插头是否连通if(c[j+2] == c[j+1]) {int tot = 0;for(int ii = 1;ii <= m+1; ii++) if(c[ii])tot++;if(tot == 2 && i == tox && j+1 == toy) ans += biao1.cnt[l];}else {if(s[i][j+1] != '.')    continue;fuzhi();for(int ii = 1;ii <= m+1; ii++) if(ii != j+1 && ii != j+2 && d[ii] == d[j+1]) {d[ii] = d[j+2];break;}d[j+1] = d[j+2] = 0;change();LL now = getval();biao2.add(now, biao1.cnt[l]);}}}biao1 = biao2;}}printf("%I64d\n", ans);}return 0;
}