洛谷P1005 矩阵取数游戏

区间dp，转移方程为

 dp[i][j] = max(dp[i-1][j]+a[k][i-1]*fac[m-j+i-1],dp[i][j]);
 dp[i][j] = max(dp[i][j+1]+a[k][j+1]*fac[m-j+i-1],dp[i][j]);

dp[i][j] 代表区[ i , j ] 的最大值，每一个状态 dp[i][j] 只能从 dp[i-1][j] 和 dp[i][j+1] 转移过来，并且取了多少次为 m-j+i-1 

最后dp[i][i] 为每个位置最后剩下的数，遍历一遍并且拿掉最后一个数即可。

重点是高精度，long long 会炸。。。（为啥不搞个 mod

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 1e3 + 10;
string add(string a,string b)//只限两个非负整数相加
{
    string ans;
    const int L = 200;
    int numa[L] = {0},numb[L] = {0};
    int lena = (int)a.size(),lenb = (int)b.size();
    for(int i=0;i<lena;i++) {
        numa[lena-1-i] = a[i]-'0';
    }
    for(int i=0;i<lenb;i++) {
        numb[lenb-1-i] = b[i]-'0';
    }
    int len = lena > lenb ? lena : lenb;
    for(int i=0;i<len;i++) {
        numa[i] += numb[i];
        numa[i+1] += numa[i]/10;
        numa[i]%=10;
    }
    if(numa[len]) len++;
    for(int i=len-1;i>=0;i--)
        ans += numa[i]+'0';
    return ans;
}
string mul(string a,string b)//高精度乘法a,b,均为非负整数
{
    string ans;
    const int L = 200;
    int numa[L],numb[L],numc[L],lena = (int)a.size(),lenb = (int)b.size();//na存储被乘数，nb存储乘数，nc存储积
    memset(numa, 0, sizeof(numa));
    memset(numb, 0, sizeof(numb));
    memset(numc, 0, sizeof(numc));
    for(int i = lena-1; i>=0; i--)
        numa[lena-i] = a[i] - '0';//将字符串表示的大整形数转成i整形数组表示的大整形数
    for(int i = lenb-1;i>=0;i--)
        numb[lenb-i] = b[i]-'0';
    for(int i=1;i<=lena;i++)
        for(int j=1;j<=lenb;j++)
            numc[i+j-1] += numa[i]*numb[j];//a的第i位乘以b的第j位为积的第i+j-1位（先不考虑进位）
    for(int i=1; i<=lena+lenb ;i++){
        numc[i+1]+=numc[i]/10;
        numc[i]%=10;//统一处理进位
    }
    if(numc[lena+lenb])
        ans += numc[lena+lenb] + '0';//判断第i+j位上的数字是不是0
    for(int i= lena + lenb - 1;i>=1;i--)
        ans += numc[i] + '0';//将整形数组转成字符串
    return ans;
}
int cmp(string a,string b){
    if (a[0] == '-' && b[0] != '-') return -1;
    if (a[0] != '-' && b[0] == '-') return 1;
    if (a[0] == '-') {
        if (a.size() > b.size()) return -1;
        else if (a.size() < b.size()) return 1;
        else{
            if (a < b) return 1;
            else if (a > b) return -1;
            else return 0;
        }
    }else{
        if (a.size() > b.size()) return 1;
        else if (a.size() < b.size()) return -1;
        else{
            if (a < b) return -1;
            else if (a > b) return 1;
            else return 0;
        }
    }
}
string fac[maxn];
string a[maxn][maxn];
string dp[maxn][maxn];
int main(){
    //freopen("/Users/chutong/data.txt", "r", stdin);
    int n,m;
    scanf("%d%d",&n,&m);
    fac[0] = "1";
    string res = "0";
    for (int i=1; i<=100; i++){
        fac[i] = mul(fac[i-1],"2");
    }
    for (int i=0; i<=n+1; i++)
        for (int j=0; j<=m+1; j++)
            dp[i][j] = a[i][j] = "0";
    for (int k=1; k<=n; k++) {
        for (int j=1; j<=m; j++)
            cin >> a[k][j];
        for(int i=0; i<=m+1; i++)
            for(int j=0; j<=m+1; j++)
                dp[i][j] = "0";
        for (int i=1; i<=m; i++) {
            for (int j=m; j>=i; j--) {
                string now1 = add(dp[i-1][j],mul(a[k][i-1],fac[m-j+i-1]));
                if (cmp(now1, dp[i][j]) == 1) dp[i][j] = now1;
                string now2 = add(dp[i][j+1],mul(a[k][j+1],fac[m-j+i-1]));
                if (cmp(now2, dp[i][j]) == 1) dp[i][j] = now2;
            }
        }
        string maxx = "0";
        for (int i=1; i<=m; i++){
            string now = add(dp[i][i], mul(a[k][i], fac[m]));
            if (cmp(now, maxx) == 1) maxx = now;
        }
        res = add(res, maxx);
    }
    cout << res << endl;
    return 0;
}