MPI Maelstrom poj 1502

来源：互联网发布：windows不是正版7601 编辑：程序博客网时间：2024/04/29 22:39

这道题算是过得顺利的，一次ac，水。

主要是处理字符串和数字的转换~

acsii码就是（字符-48）即为数字。

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>#include<cmath>using namespace std;const int inf = 0x3f3f3f3f;int edge[105][105],n,dict[105];bool vis[105];int maxx(int x,int y) {    if(x > y)         return x;    else         return y;}void dijkstra() {    int min,k,now = 1;    for(int t = 1;t <= n;t++) {        min = inf;        for(int i = 2;i <= n;i++) {            if(!vis[i]) {                if(edge[now][i] != inf && dict[i] > dict[now] + edge[now][i])                    dict[i] = dict[now] + edge[now][i];                if(min > dict[i])                    min = dict[k = i];            }        }        vis[k] = true;        now = k;    }}int main() {    while(scanf("%d",&n) != EOF) {        char chao[20];int num;        memset(vis,false,sizeof(vis));        memset(dict,inf,sizeof(dict));        dict[1] = 0;vis[1] = true;        for(int i = 2;i <= n;i++) {            for(int j = 1;j <= i-1;j++) {//输入字符串并且转换为数字                cin>>chao;                if(chao[0] == 'x')                    num = inf;                else  {                    num = 0;                    int len = strlen(chao);                    int cnt = 0,tt,nn;                    while(chao[cnt] != '\0') {                        nn = len-1,tt = 1;                        while(nn--) tt *= 10;                        num += (chao[cnt]-48)*tt;                        len--;cnt++;                    }                }                edge[i][j] = edge[j][i] = num;            }        }        for(int i = 1;i <= n;i++)            edge[i][i] = 0;        dijkstra();        int ans = 0;        for(int i = 1;i <= n;i++)             ans = maxx(ans,dict[i]);        printf("%d\n",ans);    }}

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