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几十行代码就搞定俄罗斯方块

时间:2016-08-01 15:24:55      阅读:125      评论:0      收藏:0      [点我收藏+]

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写的不错,思路也很清晰,学习一下!

 

<!doctype html>
<html>

<head>
    <title>俄罗斯方块</title>
    <style type="text/css">
      body { color:#dbdbdb; background:#000; font:25px/25px 宋体; }
      #box { position:absolute; left:50%; top:10px; margin-left:-220px; width:252px;border:#999 5px ridge; }
      #info {    position:absolute; left:50%; top:30px; margin-left:90px; font-family: Microsoft YaHei;}
      #next { padding:8px;width:105px; font-family: 宋体;}
      #rend { padding-top:30px; font:16px/25px 宋体;font-family: Microsoft YaHei; }
      #rend span { color:#ffff00; }
    </style>
</head>

<body class="c0">
    <div id="box" class="color"></div>
    <div id="info">
        NEXT:<div id="next" class="color"></div>
        <div id="text"></div>
        <div id="rend">
            <span>←↓→</span>:控制方向<br />
            <span>Z X ↑</span>:改变形状<br />
            <span>Shift</span> :一键到底<br />
            <span>空格键</span>:暂停<br />
        </div>
    </div>
    
    <script type="text/javascript">
        var map=eval("["+Array(23).join("0x801,")+"0xfff]");
        var tatris=[
            [0x6600],
            [0x2222,0xf00],
            [0xc600,0x2640],
            [0x6c00,0x4620],
            [0x4460,0x2e0,0x6220,0x740],
            [0x2260,0xe20,0x6440,0x4700],
            [0x2620,0x720,0x2320,0x2700]
        ];
        var char={x:"\u3000",s:"\u25a0",t:"\u25a1"};
        var downkey={}, 
            retkey={"40":3,"37":6,"39":6},  
            keycom={
                "38":"rotate(1)",
                "90":"rotate(1)",
                "88":"rotate(3)",
                "40":"down()","16":
                "fastdown()",
                "37":"move(1)",
                "39":"move(-1)",
                "32":"0;pause=!pause"
            };
        var dia, pos, bak, run, next, pause=false, info={speed:1,lines:0,score:0};
        function start(){
            dia=next.d;
            bak=pos={fk:[],y:0,x:4,s:next.s};
            nextdia();
            document.getElementById("next").innerHTML=(next.d[next.s%next.d.length]|0x10000).toString(2).slice(-16).replace(/..../g,"$&<br/>").replace(/1/g,char.t).replace(/0/g,char.x);
             document.getElementById("text").innerHTML="SCORE:"+info.score+"<br/><br/>LINES:"+info.lines+"<br/><br/>SPEED:"+info.speed;
            rotate(0);
            run=setInterval("pause||down()",~~(Math.pow(1.3,12-info.speed)*30+20));
        }
        function over(){
            clearInterval(kTime);
            alert("GAME OVER");
        }
        function nextdia(){
            next={d:tatris[~~(Math.random()*7)],s:~~(Math.random()*4)};
        }
        function update(t){
            t&&(pos=bak);
            bak={fk:pos.fk.slice(0),y:pos.y,x:pos.x,s:pos.s};
            if(t) return;
            for(var i=0,a2=""; i<22; i++)
                a2+=map[i].toString(2).slice(1,-1)+"<br/>";
            for(var i=0,n; i<4; i++)
                if(/([^0]+)/.test(bak.fk[i].toString(2).replace(/1/g,char.t)))
                    a2=a2.substr(0,n=(bak.y+i+1)*15-RegExp.$_.length-4)+RegExp.$1+a2.slice(n+RegExp.$1.length);
            document.getElementById("box").innerHTML=a2.replace(/1/g,char.s).replace(/0/g,char.x);
        }
        function is(){
            for(var i=0; i<4; i++)
                if((pos.fk[i]&map[pos.y+i])!=0) return true;
        }
        function rotate(r){
            var f=dia[pos.s=(pos.s+r)%dia.length];
            for(var i=0; i<4; i++)
                pos.fk[i]=(f>>(12-i*4)&15)<<pos.x;
            update(is()&&ert(1)&&ert(-2)&&ert(-1));
        }
        function down(){
            ++pos.y;
            if(is()){
                pos=bak;
                for(var i=0, r=0; i<4 && pos.y+i<22; i++)
                    if((map[pos.y+i]|=pos.fk[i])==0xfff){
                        map.splice(pos.y+i,1);
                        map.unshift(0x801);
                        r++;
                        if(++info.lines%20==0)
                            document.body.className="c"+info.speed++%7;
                    }
                clearInterval(run);
                if(map[1]!=0x801) return over();
                info.score+=~~(Math.pow(r,1.8)*10)+2;
                start();
            }
            update();
        }
        function fastdown(){
            clearInterval(run);
            run=setInterval("pause||down()",12);
        }
        function move(k){
            update(ert(k));
        }
        function ert(k){
            pos.x+=k;
            for(var i=0; i<4; i++)
                k>0 ? pos.fk[i]<<=k : pos.fk[i]>>=-k;
            return is();
        }
        document.onkeydown=function(e){
            var k=(e||event).keyCode;
            if(keycom[k] && !(k in downkey))
                downkey[k]=0;
        };
        document.onkeyup=function(e){
            var k=(e||event).keyCode;
            if(keycom[k])
                delete downkey[k];
        };
        var kTime = setInterval(function(){
            for(var k in downkey)
                if(++downkey[k]==1 || retkey[k] && downkey[k]>retkey[k])
                    eval("pause||"+keycom[k]);
        },25);
        nextdia();
        start();
    </script>

</body>

</html>

 

几十行代码就搞定俄罗斯方块

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原文地址:http://www.cnblogs.com/shoestrong/p/5725871.html

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