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            USACO 2.2 Party Lamps


            關(guān)鍵在于發(fā)現(xiàn)幾個(gè)規(guī)律:
            某一按鈕,按偶數(shù)次等同于沒(méi)按。
            按鈕的先后次序不影響結(jié)果。
            燈的狀態(tài)6位一組不變(解題時(shí)沒(méi)發(fā)現(xiàn)這個(gè)規(guī)律,開(kāi)了個(gè)100大小的bool數(shù)組)

            剩下就是幾個(gè)操作的排列組合了。代碼寫(xiě)得很繁...

            #include?<iostream>
            #include?
            <fstream>
            #include?
            <vector>
            #include?
            <algorithm>

            using?namespace?std;

            ifstream?fin(
            "lamps.in");
            ofstream?fout(
            "lamps.out");

            #ifdef?_DEBUG
            #define?out?cout
            #define?in?cin
            #else
            #define?out?fout
            #define?in?fin
            #endif


            struct?state_node{
            ????
            bool?states[101];
            ????
            int?len;
            ????
            bool?operator<(const?state_node?&?n1)?const{
            ????????
            for(int?i=1;i<=len;++i){
            ????????????
            if(states[i]!=n1.states[i])
            ????????????????
            return?states[i]<n1.states[i];
            ????????}
            ????????
            return?false;
            ????}
            ????
            bool?operator==(const?state_node?&?n1)?const{
            ????????
            for(int?i=1;i<=len;++i){
            ????????????
            if(states[i]!=n1.states[i])
            ????????????????
            return?false;
            ????????}
            ????????
            return?true;
            ????}
            };

            state_node?node;

            int?n,c;
            vector
            <int>finalon,finaloff;

            vector
            <state_node>?result;

            void?op(int?kind)
            {
            ????
            switch(kind){
            ????????
            case?1:
            ????????????
            for(int?i=1;i<=n;++i)
            ?????????????????node.states[i]
            =!node.states[i];
            ????????????
            break;
            ????????
            case?2:
            ????????????
            for(int?i=1;i<=n;++i){
            ????????????????
            if(i&1)
            ????????????????????node.states[i]
            =!node.states[i];
            ????????????}
            ????????????
            break;
            ????????
            case?3:
            ????????????
            for(int?i=1;i<=n;++i){
            ????????????????
            if(!(i&1))
            ????????????????????node.states[i]
            =!node.states[i];
            ????????????}
            ????????????
            break;
            ????????
            case?4:
            ????????????
            for(int?i=1;i<=n;++i){
            ????????????????
            if(i%3==1){
            ????????????????????node.states[i]
            =!node.states[i];
            ????????????????}
            ????????????}
            ????????????
            break;
            ????}
            }

            bool?isok()
            {
            ????
            for(int?i=0;i<finalon.size();++i){
            ????????
            if(!node.states[finalon[i]])
            ????????????
            return?false;
            ????}
            ????
            for(int?i=0;i<finaloff.size();++i){
            ????????
            if(node.states[finaloff[i]])
            ????????????
            return?false;
            ????}
            ????
            return?true;
            }

            void?do0()
            {
            ????
            if(isok())
            ????????result.push_back(node);
            }

            void?do1()
            {
            ????
            for(int?i=1;i<=4;++i){
            ????????op(i);
            ????????
            if(isok()){
            ????????????result.push_back(node);
            ????????}
            ????????op(i);
            ????}
            }

            void?do2()
            {
            ????
            for(int?i=1;i<=3;++i){
            ????????
            for(int?j=i+1;j<=4;++j){
            ????????????op(i);
            ????????????op(j);
            ????????????
            if(isok()){
            ????????????????result.push_back(node);
            ????????????}
            ????????????op(i);
            ????????????op(j);

            ????????}
            ????}
            }

            void?do3()
            {
            ????
            //操作1,2,3,4各執(zhí)行一次,前三個(gè)操作抵消
            ????op(4);
            ????do1();
            }

            void?do4()
            {
            ????op(
            4);
            ????
            if(isok()){
            ????????result.push_back(node);
            ????}
            ????op(
            4);
            }

            void?action(int?i)
            {
            ????
            switch(i){
            ????????
            case?0:?do0();break;
            ????????
            case?1:?do1();break;
            ????????
            case?2:?do2();break;
            ????????
            case?3:?do3();break;
            ????????
            case?4:?do4();break;
            ????}
            }

            void?solve()
            {
            ????
            in>>n>>c;

            ????node.len?
            =?n;

            ????
            for(int?i=1;i<=n;++i)?
            ????????node.states[i]?
            =?true;

            ????
            int?t;
            ????
            do{
            ????????
            in>>t;
            ????????
            if(t!=-1)
            ????????????finalon.push_back(t);
            ????}
            while(t!=-1);

            ?????
            do{
            ????????
            in>>t;
            ????????
            if(t!=-1)
            ????????????finaloff.push_back(t);
            ????}
            while(t!=-1);

            ????
            bool?has_result?=?false;

            ????
            for(int?i=0;i<=4;++i){
            ????????
            if(c>=i&&(c-i)%2==0){
            ????????????action(i);
            ????????}
            ????}

            ????sort(result.begin(),result.end());
            ????vector
            <state_node>::iterator?end?=?
            ????????unique(result.begin(),result.end());

            ????
            for(vector<state_node>::iterator?i?=?result.begin();
            ????????????i
            !=end;?++i){
            ????????has_result?
            =?true;
            ????????
            for(int?j=1;j<=n;++j){
            ????????????
            out<<i->states[j];
            ????????}
            ????????
            out<<endl;
            ????}

            ????
            if(!has_result){
            ????????
            out<<"IMPOSSIBLE"<<endl;
            ????}
            }

            int?main(int?argc,char?*argv[])
            {
            ????solve();?
            ????
            return?0;
            }




            posted on 2009-06-20 20:33 YZY 閱讀(1343) 評(píng)論(0)  編輯 收藏 引用 所屬分類(lèi): AlgorithmUSACO

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