（intermediate）最短路（最长路） UVA 1348 - Tomato Automata

1348 - Tomato Automata

Time limit: 3.000 seconds

Tomato Automata are small cool programs. You give them an infinite sequence of ones and zeros, and they give you a sequence of numbers. They are widely used in the Stanescu Operating System (SOS). They are written in Tomato Programming Language that is very simple. Here is its specification, tutorial and reference:

1. Tomato is a very simple but powerful language.
2. Lines in a Tomato program are numerated with integers from 1 to N (N100000) in the order they appear in the input.
3. There is exactly one command on a line.
4. The execution starts from the first line.
5. There are exactly five Tomato commands - ifgo, jump, pass, loop, and die.
6. When executed, each command prints its line number (into the output sequence).
7. The ifgo command has one argument - the line number of another instruction. It reads one bit from the input stream. If the bit is one, the execution jumps to the line with the given as argument line number. Otherwise the execution continues with the next line.
8. The jump command has one argument - the line number of another instruction. When executed, the execution jumps to the line with the given line number.
9. The pass command has no arguments. It does nothing (except printing its line number like all other commands). Then the execution continues with the next line.
10. The die command has no arguments. It terminates the execution of the program (printing its line number before that). The die command can not be used inside a loop.
11. The loop command is the only one with two arguments. It may be used to construct loops. The first argument is the starting line number < line > (less than the line number of the loop command), and the second is a positive integer < count > . When executed, it loops from the start line a < count > - 1 number of times (because it is already executed once). When the loop is executed the given number of times, the execution continues with the next line.
12. The jump and ifgo commands must be used only with line numbers in the scope of the innermost loop containing them (they can not jump outside of the innermost loop or inside loops nested in the innermost loop that does not contain the command).
13. The loop command can not be used to create overlapping loops. Nested loops must be strictly nested (they can not use the same starting line).
14. When the last line of the program is executed, the execution continues from the first except when the last line is die command.
15. White spaces may occur freely before or after the command name and their arguments.
16. The maximal length of a line in Tomato Programming Language is 80 characters including spaces.

Stanescu has lots of Tomato programs. He is interested in maximal length of output sequence that specific program can generate, where the length is the number of printed line numbers. Obviously, it is not possible to test each possible input sequence (of ones and zeros), so he needs a program that computes this.

Input 

The input contains several programs, separated with an empty line. Each of them is a correct Tomato program.

Output 

For each given program your solution has to print on a separate line the maximal length ofthe output sequence the program could generate. Print `infinity' if there is no maximal length for the 9 output sequence. When finite, the maximal length will not exceed 10⁹ .

Sample Input 

Ifgo 2 loop 1 3 die ifgo 2 ifgo 3 pass die       pass             ifgo 4             jump 5             ifgo 3       loop 2 2       pass loop 1 2 die

Sample Output 

7 4  

23 

题意：基本上就是从第一句开始最多能走多少步，如果能走进死循环就是infinite，否则就输出最多的步数。

思路：这个做到我吐血啊，poj过不了啊啊啊啊，uva上过了，因为时间宽松一点。这题麻烦的地方时建图，呵呵，真的会吐血！！首先第一句是一个点，所有if语句是一个点，loop的语句和它跳到的语句是点，die也是点，开始建图吧，如果建图的过程中有死循环，就加一条自环，最后的话spfa。。。。我想过可以把图转化为dag，但是好麻烦啊，如果转换为了dag，发现从起点开始能进入一个不是一个点的强连通的话，就是infinite，如果不能，就先拓扑排序，再用动态规划做。但是我真的写不动了，要写好长，好麻烦。。。

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<string.h>#include<vector>#include<queue>#include<stack>using namespace std;#define LL long longconst int maxn = 100000+5;const int inf = 1e9;int n , m , ans , size;LL d[maxn] , cnt[maxn] , succ[maxn] , len[maxn];int dfn , pre[maxn] , low[maxn];bool sccno[maxn];stack<int> S;int sign[maxn];int pos[maxn];bool inq[maxn] , die[maxn];char cmd[maxn][82];char buffer[82];struct Node{int v , w , pre , t;bool loop;Node *next;}edge[2*maxn] , *first[maxn];void add(int x,int y,int w,int pre,int t,bool loop){edge[++m].v = y; edge[m].w = w;edge[m].pre = pre; edge[m].t = t;edge[m].loop = loop;edge[m].next = first[x];first[x] = &edge[m];}void init(){m = 0;while (S.size()) S.pop();memset(pre,0,sizeof(pre));memset(sccno,false,sizeof(sccno));dfn = 0;memset(first,0,sizeof(first));}bool cmd_modify(){bool die = false;for (int i = 0 ; i < n ; ++i){if (cmd[i][0]=='d') die = true;if (cmd[i][0]=='i' || cmd[i][0]=='j') {int v;sscanf(cmd[i],"%*s%d",&v);--v;if ((i+1)%n==v) cmd[i][0] = 'p';if (i==v && !die) return false;}}return die;}bool dfs(int u){pre[u] = low[u] = ++dfn;S.push(u);for (Node * p = first[u] ; p ; p=p->next){int v = p->v;if (!pre[v]){if (!dfs(v)) return false;low[u] = min(low[u],low[v]);} else if (!sccno[v])low[u] = min(low[u],pre[v]);}if (pre[u]==low[u]) {if (u==0 && S.size() > 1) return false;while (true) {int x = S.top() ; S.pop();sccno[x] = true;if (x==u) break;}}return true;}void BuildGraph(){memset(sign,-1,sizeof(sign));memset(pos,-1,sizeof(pos));memset(die,0,sizeof(die));sign[0] = ++size;pos[size] = 0;int u , v , w , pre , t;for (int i = 0 ; i < n ; ++i){if (cmd[i][0]=='l') {sscanf(cmd[i],"%*s%d%d",&pre,&t);--pre;if (sign[pre]==-1) { sign[pre] = ++size; pos[size] = pre; }if (sign[i]==-1) { sign[i] = ++size; pos[size] = i; }succ[i] = i;} else if (cmd[i][0]=='i' || cmd[i][0]=='j') {sscanf(cmd[i],"%*s%d",&v);--v;if (cmd[i][0]=='i' && sign[i]==-1) { sign[i] = ++size; pos[size] = i; }succ[i] = v;} else if (cmd[i][0]=='d' ) {if (sign[i]==-1) { sign[i] = ++size; pos[size] = i; }die[sign[i]] = true;succ[i] = i;} else  { succ[i] = (i+1)%n; }len[i] = succ[i]==i ? 0 : 1;}for (int i = size ; i >=1 ; --i){if (die[i]) continue;int u = pos[i] ,  v , w , pre = -1 , t = -1;v = succ[u] , w = len[u];memset(inq,false,sizeof(inq));inq[u] = true;bool lp = false;while (sign[v]==-1){inq[v] = true;w += len[v];v = succ[v];if (inq[v]) { lp = true; break; }}if (lp) { add(sign[u],sign[u],inf,-1,-1,false); continue; }len[u] = w;succ[u] = v;if (v!=u) { add(sign[u],sign[v],w,-1,-1,false); }if (cmd[u][0]=='j') continue;v = (u+1)%n , w = 1;memset(inq,false,sizeof(inq));lp = false;inq[u] = true;while (sign[v]==-1) {inq[v] = true;w += len[v];v = succ[v];if (inq[v]) { lp = true; break; }}if (lp) { add(sign[u],sign[u],inf,-1,-1,false); continue; }if (cmd[u][0]=='l') sscanf(cmd[u],"%*s%d%d",&pre,&t);--pre;add(sign[u],sign[v],w,sign[pre],t,cmd[u][0]=='l');}}void spfa(){memset(cnt,0,sizeof(cnt));memset(inq,0,sizeof(inq));memset(d,0,sizeof(d));queue<int> q;q.push(sign[0]);++cnt[sign[0]];d[sign[0]] = 1;int u , v , w;while (q.size()){u = q.front(); q.pop();inq[u] = false;if (die[u] && ans < d[u]) ans = d[u];for (Node * p = first[u] ; p ; p = p->next){int v = p->v , w = p->w;if (p->loop) w += (d[u]-d[p->pre]+1)*(p->t-1);if (d[v] < d[u]+w){d[v] = d[u]+w;if (!inq[v]){inq[v] = true; q.push(v);++cnt[v];if (cnt[v] >= size) {printf("infinity\n");return;}}}}}cout << ans << endl;}void get_cmd(){for (int i = 0 ; i < strlen(buffer) ; ++i)buffer[i] = tolower(buffer[i]);char * p = strtok(buffer," ");while (p) {strcat(cmd[n],p);strcat(cmd[n]," ");p = strtok(NULL," ");}++n;}int main(){//freopen("input.in","r",stdin);while (gets(buffer)) {if (buffer[0]==0) break;memset(cmd,0,sizeof(cmd));n = size = 0;get_cmd();while (gets(buffer)) {if (buffer[0]==0) break;for (int i = 0 ; i < strlen(buffer) ; ++i)if (isspace(buffer[i])) buffer[i] = ' ';get_cmd();}if (!cmd_modify()) { printf("infinity\n"); continue; }init();BuildGraph();if (!dfs(0)) { printf("infinity\n"); continue; }ans = -1;spfa();}}