將每個(gè)translation的輸入和輸出place全部記錄下來(lái)，模擬即可，當(dāng)所有translation都不能工作時(shí)，就說(shuō)明dead了。

AC代碼：

#include<cstdio>#include<vector>using namespace std;const int maxn = 100 + 5;struct node{    vector<int> in, out;}tran[maxn];int p[maxn]; //the number of tokens in  all placesint main(){    int pn, tn, nf, kase = 1;    while(scanf("%d", &pn) == 1 && pn){        for(int i = 1; i <= pn; ++i){            scanf("%d", &p[i]);        }        scanf("%d", &tn);        for(int i = 1; i <= tn; ++i){            int x;            while(scanf("%d", &x) == 1 && x){                if(x < 0) tran[i].in.push_back(-x);                else tran[i].out.push_back(x);            }        }        scanf("%d",&nf);        bool dead = 0;        int h;        for(h = 0; h < nf; ++h){            int cnt = 0;            for(int i = 1; i <= tn; ++i){                bool flag = 1;                vector<int> &in = tran[i].in, &out = tran[i].out;                for(int j = 0; j < in.size(); ++j){                    if(p[in[j]] == 0) {                        flag = 0;                        while(j) p[in[--j]]++; //                        break;                    }                    else p[in[j]]--;                }                if(!flag) ++cnt;                else {                    for(int k = 0; k < out.size(); ++k) p[out[k]]++;                    break;                }            }            if(cnt == tn) {               dead = 1;               break;            }        }        if(dead) PRintf("Case %d: dead after %d transitions/n", kase++, h);        else printf("Case %d: still live after %d transitions/n", kase++, nf);        printf("Places with tokens:");        for(int i = 1; i <= pn ; ++i){            if(p[i]) printf(" %d (%d)", i, p[i]);        }        printf("/n/n");        for(int i = 1; i <= tn ; ++i) {            tran[i].in.clear();            tran[i].out.clear();        }    }    return 0;}如有不當(dāng)之處歡迎指出！