題意：給一個元素周期表的元素符號（114種），再給一個串，問這個串能否有這些元素符號組成（全為小寫）。

解法1：動態規劃

定義：dp[i]表示到 i 這個字符為止，能否有元素周期表里的符號構成。

則有轉移方程：dp[i] = (dp[i-1]&&f(i-1,1)) || (dp[i-2]&&f(i-2,2)) ? ? f(i,k):表示從i開始填入k個字符，這k個字符在不在元素周期表中。 ?dp[0] = 1

代碼：

//109ms 0KB #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <string> using namespace std; #define N 50007string single[25] = {"h","b","c","n","o","f","k","p","s","y","i","w","u","v"}; string ss[130] = {"he","li","be","ne","na","mg", "al","si","cl","ar","ca","sc","ti","cr","mn", "fe","co","ni","cu","zn","ga","ge","as","se", "br","kr","rb","sr","zr","nb","mo","tc","ru", "rh","pd","ag","cd","in","sn","sb","te","xe", "cs","ba","hf","ta","re","os","ir","pt","au", "hg","tl","pb","bi","po","at","rn","fr","ra", "rf","db","sg","bh","hs","mt","ds","rg","cn", "fl","lv","la","ce","pr","nd","pm","sm","eu", "gd","tb","dy","ho","er","tm","yb","lu","ac", "th","pa","np","pu","am","cm","bk","cf","es", "fm","md","no","lr"};int vis[30][30],tag[30]; int dp[N]; char st[N];void init() {memset(vis,0,sizeof(vis));memset(tag,0,sizeof(tag));for(int i=0;i<14;i++)tag[single[i][0]-'a'] = 1;for(int i=0;i<100;i++)vis[ss[i][0]-'a'][ss[i][1]-'a'] = 1; }int main() {int t,len,i;init();scanf("%d",&t);while(t--){scanf("%s",st+1);len = strlen(st+1);memset(dp,0,sizeof(dp));dp[0] = 1;for(i=0;i<len;i++){if(dp[i]){if(tag[st[i+1]-'a'])dp[i+1] = 1;dp[i+2] |= vis[st[i+1]-'a'][st[i+2]-'a'];}}if(dp[len])puts("YES");elseputs("NO");}return 0; } View Code

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解法2：DFS

搜索時循環的是元素周期表的符號個數。詳見代碼

代碼： （306ms）

#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <string> using namespace std; #define N 50007string ss[130] = {"h","b","c","n","o","f","k","p","s","y","i","w","u","v","he","li","be","ne","na","mg", "al","si","cl","ar","ca","sc","ti","cr","mn", "fe","co","ni","cu","zn","ga","ge","as","se", "br","kr","rb","sr","zr","nb","mo","tc","ru", "rh","pd","ag","cd","in","sn","sb","te","xe", "cs","ba","hf","ta","re","os","ir","pt","au", "hg","tl","pb","bi","po","at","rn","fr","ra", "rf","db","sg","bh","hs","mt","ds","rg","cn", "fl","lv","la","ce","pr","nd","pm","sm","eu", "gd","tb","dy","ho","er","tm","yb","lu","ac", "th","pa","np","pu","am","cm","bk","cf","es", "fm","md","no","lr"};int vis[N]; int len[140]; char st[N]; int Length; bool Tag;void init() {int i;for(i=0;i<14;i++)len[i] = 1;for(i=14;i<114;i++)len[i] = 2; }void dfs(int u) {if(u == Length)Tag = 1;if(Tag)return;for(int i=0;i<114;i++){int flag = 1;if(u+len[i] <= Length && !vis[u+len[i]]){for(int j=0;j<len[i];j++){if(ss[i][j] != st[u+j]){flag = 0;break;}}if(flag){vis[u+len[i]] = 1;dfs(u+len[i]);}}} }int main() {init();int t,i;scanf("%d",&t);while(t--){scanf("%s",st);Length = strlen(st);memset(vis,0,sizeof(vis));Tag = 0;dfs(0);if(Tag)puts("YES");elseputs("NO");}return 0; } View Code

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解法3：亂搞，模擬。

分成： 單個元素存在與否，與前面匹不匹配，與后面匹不匹配，總共2^3 = 8種情況，然后O(n)掃過去，代碼很長。。。

代碼：（586ms）

#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <string> using namespace std; #define N 50007string single[25] = {"h","b","c","n","o","f","k","p","s","y","i","w","u","v"}; string ss[130] = {"he","li","be","ne","na","mg", "al","si","cl","ar","ca","sc","ti","cr","mn", "fe","co","ni","cu","zn","ga","ge","as","se", "br","kr","rb","sr","zr","nb","mo","tc","ru", "rh","pd","ag","cd","in","sn","sb","te","xe", "cs","ba","hf","ta","re","os","ir","pt","au", "hg","tl","pb","bi","po","at","rn","fr","ra", "rf","db","sg","bh","hs","mt","ds","rg","cn", "fl","lv","la","ce","pr","nd","pm","sm","eu", "gd","tb","dy","ho","er","tm","yb","lu","ac", "th","pa","np","pu","am","cm","bk","cf","es", "fm","md","no","lr"};char st[N]; int vis[N];int main() {int t,len,i,j,k;scanf("%d",&t);while(t--){scanf("%s",st);len = strlen(st);int flag = 1;memset(vis,0,sizeof(vis));for(i=0;i<len;i++){if(vis[i])continue;string S = "";S += st[i];for(j=0;j<14;j++){if(single[j] == S)break;}if(j == 14) //not single {if(i > 0 && !vis[i-1]){S = st[i-1]+S;for(j=0;j<100;j++){if(ss[j] == S)break;}if(j != 100) //pre match {if(i < len-1){string ks = "";ks += st[i];ks += st[i+1];for(k=0;k<100;k++){if(ss[k] == ks)break;}if(k != 100) //back matchvis[i] = 0;else //back not matchvis[i] = 1;}}else //pre not match {if(i < len-1){string ks = "";ks += st[i];ks += st[i+1];for(k=0;k<100;k++){if(ss[k] == ks)break;}if(k != 100) //back matchvis[i+1] = 1;else //back not match {flag = 0;break;}}else{flag = 0;break;}}}else{if(i < len-1){string ks = "";ks += st[i];ks += st[i+1];for(k=0;k<100;k++){if(ss[k] == ks)break;}if(k != 100) //back matchvis[i+1] = 1;else //back not match {flag = 0;break;}}else{flag = 0;break;}}}else //single {if(i > 0 && !vis[i-1]){S = st[i-1]+S;for(j=0;j<100;j++){if(ss[j] == S)break;}if(j != 100) //pre match {if(i < len-1){string ks = "";ks += st[i];ks += st[i+1];for(k=0;k<100;k++){if(ss[k] == ks)break;}if(k != 100) //back matchvis[i] = 0;else //back not matchvis[i] = 1;}}else //pre not match {if(i < len-1){string ks = "";ks += st[i];ks += st[i+1];for(k=0;k<100;k++){if(ss[k] == ks)break;}if(k != 100) //back matchvis[i] = 0;else //back not matchvis[i] = 1;}elsevis[i] = 1;}}else{if(i < len-1){string ks = "";ks += st[i];ks += st[i+1];for(k=0;k<100;k++){if(ss[k] == ks)break;}if(k != 100) //back matchvis[i] = 0;else //back not matchvis[i] = 1;}elsevis[i] = 1;}}}if(flag)puts("YES");elseputs("NO");}return 0; } View Code

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轉載于:https://www.cnblogs.com/whatbeg/p/3876636.html

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