Codeforces Round #106 (Div. 2) D. Coloring Brackets 区间dp

题目链接：

http://codeforces.com/problemset/problem/149/D

D. Coloring Brackets

time limit per test2 seconds
memory limit per test256 megabytes
#### 问题描述 > Once Petya read a problem about a bracket sequence. He gave it much thought but didn't find a solution. Today you will face it. > > You are given string s. It represents a correct bracket sequence. A correct bracket sequence is the sequence of opening ("(") and closing (")") brackets, such that it is possible to obtain a correct mathematical expression from it, inserting numbers and operators between the brackets. For example, such sequences as "(())()" and "()" are correct bracket sequences and such sequences as ")()" and "(()" are not. > > In a correct bracket sequence each bracket corresponds to the matching bracket (an opening bracket corresponds to the matching closing bracket and vice versa). For example, in a bracket sequence shown of the figure below, the third bracket corresponds to the matching sixth one and the fifth bracket corresponds to the fourth one. > > > You are allowed to color some brackets in the bracket sequence so as all three conditions are fulfilled: > > Each bracket is either not colored any color, or is colored red, or is colored blue. > For any pair of matching brackets exactly one of them is colored. In other words, for any bracket the following is true: either it or the matching bracket that corresponds to it is colored. > No two neighboring colored brackets have the same color. > Find the number of different ways to color the bracket sequence. The ways should meet the above-given conditions. Two ways of coloring are considered different if they differ in the color of at least one bracket. As the result can be quite large, print it modulo 1000000007 (109 + 7). #### 输入 > The first line contains the single string s (2 ≤ |s| ≤ 700) which represents a correct bracket sequence. #### 输出 > Print the only number — the number of ways to color the bracket sequence that meet the above given conditions modulo 1000000007 (109 + 7). ####样例输入 > (()()) ####样例输出 > 40 ## 题意 > 每对括号必须满足一边染成红色或蓝色，另一边不染色，且相邻的两个括号颜色不同。

题解

之前的一种思路是dp[l][r][s1][s2]代表最左边的括号的状态为s1，s2但是，这种是错的额！因为你看不到r右边的限制了！！！！
贴个错误代码：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=777;
const int mod=1e9+7;

LL dp[maxn][maxn][3][3];
char str[maxn];
int rig[maxn];
int n;

LL dfs(int l,int r,int s1,int s2) {
    if(l>=r) return 0;
    if(l+1==r&&(s1==0&&s2>0||s1>0||s2==0)) return 1;
    if(dp[l][r][s1][s2]>=0) return dp[l][r][s1][s2];

    LL &res=dp[l][r][s1][s2]=0;

    int ll=l,rr=rig[l];

    LL cntl=0,cntr=0;
    if(s1==0) {
        if(ll+1<rr-1) {
            cntl+=dfs(ll+1,rr-1,1,0);
            cntl+=dfs(ll+1,rr-1,2,0);
            cntl+=dfs(ll+1,rr-1,0,3-s2);
        } else {
            cntl=1;
        }
        cntl%=mod;

        if(rr+1>r) {
            cntr=1;
        } else {
            cntr+=dfs(rr+1,r,0,1);
            cntr+=dfs(rr+1,r,0,2);
            cntr+=dfs(rr+1,r,3-s2,0);
        }
        cntr%=mod;

        res+=cntl*cntr;
        res%=mod;
    } else {
        if(ll+1<rr-1) {
            cntl+=dfs(ll+1,rr-1,3-s1,0);
            cntl+=dfs(ll+1,rr-1,0,1);
            cntl+=dfs(ll+1,rr-1,0,2);
        } else {
            cntl=1;
        }
        cntl%=mod;

        if(rr+1>r) {
            cntr=1;
        } else {
            cntr+=dfs(rr+1,r,0,1);
            cntr+=dfs(rr+1,r,0,2);
            cntr+=dfs(rr+1,r,1,0);
            cntr+=dfs(rr+1,r,2,0);
        }
        cntr%=mod;

        res+=cntl*cntr;
        res%=mod;
    }

    return res;
}

int main() {
    scf("%s",str+1);
    n=strlen(str+1);
    stack<int> mst;
    clr(dp,-1);
    clr(rig,-1);
    for(int i=1; i<=n; i++) {
        if(str[i]==')') {
            rig[mst.top()]=i;
            mst.pop();
        } else {
            mst.push(i);
        }
    }
//    for(int i=1;i<=n;i++){
//        prf("(%d,%d)\n",i,rig[i]);
//    }

    LL ans=dfs(1,n,0,1)+dfs(1,n,0,2)+dfs(1,n,1,0)+dfs(1,n,2,0);
    bug(dp[1][n][0][1]);
    bug(dp[1][n][0][2]);
    bug(dp[1][n][1][0]);
    bug(dp[1][n][2][0]);

    bug(dp[2][5][2][0]);

    prf("%I64d\n",ans);

    return 0;
}

//end-----------------------------------------------------------------------

/*
(()())
*/

正确的表示应该是dp[l][r][s1][s2]代表l左边和r右边的限制！！！

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=777;
const int mod=1e9+7;

LL dp[maxn][maxn][3][3];
char str[maxn];
int rig[maxn];
int n;

LL dfs(int l,int r,int s1,int s2) {
    if(l>r) return 1;
    if(dp[l][r][s1][s2]>=0) return dp[l][r][s1][s2];

    LL &res=dp[l][r][s1][s2]=0;
    int ll=l,rr=rig[l];
    if(rr==r){
        if(s1==0){
            if(s2==0){
                res+=dfs(ll+1,rr-1,0,1);
                res+=dfs(ll+1,rr-1,0,2);
                res+=dfs(ll+1,rr-1,1,0);
                res+=dfs(ll+1,rr-1,2,0);
                res%=mod;
            }else{
                res+=dfs(ll+1,rr-1,0,3-s2);
                res+=dfs(ll+1,rr-1,1,0);
                res+=dfs(ll+1,rr-1,2,0);
                res%=mod;
            }
        }else{
            if(s2==0){
                res+=dfs(ll+1,rr-1,0,1);
                res+=dfs(ll+1,rr-1,0,2);
                res+=dfs(ll+1,rr-1,3-s1,0);
                res%=mod;
            }else{
                res+=dfs(ll+1,rr-1,0,3-s2);
                res+=dfs(ll+1,rr-1,3-s1,0);
                res%=mod;
            }
        }
    }else{
        if(s1==0){
            res=(res+dfs(ll+1,rr-1,0,1)*dfs(rr+1,r,1,s2)%mod)%mod;
            res=(res+dfs(ll+1,rr-1,0,2)*dfs(rr+1,r,2,s2)%mod)%mod;
            res=(res+dfs(ll+1,rr-1,1,0)*dfs(rr+1,r,0,s2)%mod)%mod;
            res=(res+dfs(ll+1,rr-1,2,0)*dfs(rr+1,r,0,s2)%mod)%mod;
        }else{
            res=(res+dfs(ll+1,rr-1,0,1)*dfs(rr+1,r,1,s2)%mod)%mod;
            res=(res+dfs(ll+1,rr-1,0,2)*dfs(rr+1,r,2,s2)%mod)%mod;
            res=(res+dfs(ll+1,rr-1,3-s1,0)*dfs(rr+1,r,0,s2)%mod)%mod;
        }
    }

    return res;
}

int main() {
    scf("%s",str+1);
    n=strlen(str+1);
    stack<int> mst;
    clr(dp,-1);
    clr(rig,-1);
    for(int i=1; i<=n; i++) {
        if(str[i]==')') {
            rig[mst.top()]=i;
            mst.pop();
        } else {
            mst.push(i);
        }
    }

    LL ans=dfs(1,n,0,0);
    prf("%I64d\n",ans);

    return 0;
}

//end-----------------------------------------------------------------------

/*
(()())
*/

短一点的代码：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=777;
const int mod=1e9+7;

LL dp[maxn][maxn][3][3];
char str[maxn];
int rig[maxn];
int n;

LL dfs(int l,int r,int s1,int s2) {
    if(l>r) return 1;
    if(dp[l][r][s1][s2]>=0) return dp[l][r][s1][s2];

    LL &res=dp[l][r][s1][s2]=0;
    int ll=l,rr=rig[l];
    for(int i=1;i<=2;i++){
        //left
        if(i!=s1) res=(res+dfs(ll+1,rr-1,i,0)*dfs(rr+1,r,0,s2)%mod)%mod;
        //right
        if(rr<r||i!=s2) res=(res+dfs(ll+1,rr-1,0,i)*dfs(rr+1,r,i,s2)%mod)%mod;
    }

    return res;
}

int main() {
    scf("%s",str+1);
    n=strlen(str+1);
    stack<int> mst;
    clr(dp,-1);
    clr(rig,-1);
    for(int i=1; i<=n; i++) {
        if(str[i]==')') {
            rig[mst.top()]=i;
            mst.pop();
        } else {
            mst.push(i);
        }
    }

    LL ans=dfs(1,n,0,0);
    prf("%I64d\n",ans);

    return 0;
}

//end-----------------------------------------------------------------------

/*
(()())
*/
posted @ 2016-09-28 02:37  fenicnn  阅读(186)  评论(0)  编辑  收藏  举报