UVALive - 4260 Fortune Card Game （DP&状态转移）好题

UVALive - 4260

Fortune Card Game

Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

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Description

A popular card game called ``fortune" is getting popular in country X. Fig. 1 shows one of the cards. In each card, a positive integer number (20 in the figure) is listed as the address of the card. A symbol is drawn beside the address. There are five kinds of symbols, which are listed below the card. For convenience, let each symbol be represented by an English letter from `A'-`E'. The bottom of a card contains another number called ``next fortune number."

Figure 1: A sample fortune card and symbols.

In a set of fortune cards, many cards can have same address; that is, address 20 is not limited to appear only in one card. However, there will be no cards that are identical, i.e., no two cards with same address, symbol, and next fortune number.

The fortune card game is played as follows. A player starts with cards that have address 1. The goal of the game is trying to complete a ``spell" that is composed by the symbols. For example, let a spell be ``BADDAD". In the first move, the player will look for cards that have address 1 with a star symbol (which matches `B' in the spell). The next fortune numbers of these cards are the new addresses for the next move. The player can select one card to advance to a new address x. The selected card is then put back to the cards for next move but the fortune number is written down.

Let the example be continued. In the next move, the player needs to look for the cards that have new address x with the cross symbol (which matches the second `A' in the spell). Again, the player selects one card to advance to another new address. This procedure continues until the spell ``BADDAD" is completed. Once the player completes a spell, he wins a score by adding all the next fortune numbers of the selected card, which have been written down.

Given a spell and a set of fortune cards, please output the maximum score that can be played in this card game.

Technical Specification

1. N - the number of test cases, N10 .
2. C - the number of cards, C800 .
3. L - the length of a spell, L150 .

Input

Test data begins with an integer N which is the number of test cases. Each test case begins with an integerC , which is the number of cards. Following the numberC is C lines of card information. Each card is represented by (Address Symbol NextF ortuneNumber). The address and next fortune number are between 1 and 800. The symbols are capital letters from ` A' to ` E'. The last line of a test case is a spell. The spell is a string composed by capital letters from `A' to ` E'. The length of the spell ( L ) is less than 150.

Output

For each test case, please output the maximum score that can be collected for each test case.

Sample Input

2 7 1 A 2 1 A 3 2 A 3 2 B 4 2 B 5 3 A 3 3 B 4 AAAAB 6 1 A 2 1 B 2 1 A 3 1 B 3 2 A 3 2 B 3 AB

Sample Output

16 5

//题意：

给你n张卡片，每张卡片有三个属性，ai 卡片的索引，ci 卡片的符号，bi 卡片的价值，现在给你一串字符串，让你根据这个字符串求出这个字符串的最大价值，找最大价值是有要求的：

你所拿的这张卡片的价值应为下一张卡片的索引值。

//思路：

运用动态规划求最值，dp[i][j]表示第i个字符索引值为j的最大价值那么可以得到状态转移方程为：dp[i][p[j].b]=max(dp[i][p[j].b,dp[i-1][p[j].a]+p[j].b);

有了上面的状态转移方程就可以计算了。

#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>#define INF 0x3f3f3f3f#define N 810using namespace std;struct zz{int a;char c;int b;}p[N];int dp[N][N];char s[210];int main(){int t,n;int i,j;scanf("%d",&t);while(t--){scanf("%d",&n);for(i=1;i<=n;i++)scanf("%d %c %d",&p[i].a,&p[i].c,&p[i].b);scanf("%s",s);int len=strlen(s);memset(dp,-INF,sizeof(dp));dp[0][1]=0;for(i=1;i<=len;i++){for(j=1;j<=n;j++){if(s[i-1]==p[j].c){dp[i][p[j].b]=max(dp[i][p[j].b],dp[i-1][p[j].a]+p[j].b);}}}int ans=0;for(i=1;i<=n;i++){ans=max(ans,dp[len][i]);}printf("%d\n",ans);}return 0;}